From 4b575a8dab798fc213c56d21891e6d6e7446bbc5 Mon Sep 17 00:00:00 2001 From: fw_qaq <82551626+fwqaaq@users.noreply.github.com> Date: Wed, 7 Dec 2022 23:18:23 +0800 Subject: [PATCH 001/154] =?UTF-8?q?Update=200450.=E5=88=A0=E9=99=A4?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84?= =?UTF-8?q?=E8=8A=82=E7=82=B9.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0450.删除二叉搜索树中的节点.md | 33 +++++++++++++++++++ 1 file changed, 33 insertions(+) diff --git a/problems/0450.删除二叉搜索树中的节点.md b/problems/0450.删除二叉搜索树中的节点.md index 0b4048d5..3d9a6050 100644 --- a/problems/0450.删除二叉搜索树中的节点.md +++ b/problems/0450.删除二叉搜索树中的节点.md @@ -657,6 +657,39 @@ object Solution { } ``` +## rust + +```rust +impl Solution { + pub fn delete_node( + root: Option>>, + key: i32, + ) -> Option>> { + root.as_ref()?; + + let mut node = root.as_ref().unwrap().borrow_mut(); + match node.val.cmp(&key) { + std::cmp::Ordering::Less => node.right = Self::delete_node(node.right.clone(), key), + std::cmp::Ordering::Equal => match (node.left.clone(), node.right.clone()) { + (None, None) => return None, + (None, Some(r)) => return Some(r), + (Some(l), None) => return Some(l), + (Some(l), Some(r)) => { + let mut cur = Some(r.clone()); + while let Some(n) = cur.clone().unwrap().borrow().left.clone() { + cur = Some(n); + } + cur.unwrap().borrow_mut().left = Some(l); + return Some(r); + } + }, + std::cmp::Ordering::Greater => node.left = Self::delete_node(node.left.clone(), key), + } + root.clone() + } +} +``` +

From d97f27621c96972c11b0e48b5498c411ac1e85f5 Mon Sep 17 00:00:00 2001 From: fw_qaq <82551626+fwqaaq@users.noreply.github.com> Date: Fri, 9 Dec 2022 15:44:18 +0800 Subject: [PATCH 002/154] =?UTF-8?q?Update=20problems/0450.=E5=88=A0?= =?UTF-8?q?=E9=99=A4=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD?= =?UTF-8?q?=E7=9A=84=E8=8A=82=E7=82=B9.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0450.删除二叉搜索树中的节点.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/problems/0450.删除二叉搜索树中的节点.md b/problems/0450.删除二叉搜索树中的节点.md index 3d9a6050..424491b7 100644 --- a/problems/0450.删除二叉搜索树中的节点.md +++ b/problems/0450.删除二叉搜索树中的节点.md @@ -685,7 +685,8 @@ impl Solution { }, std::cmp::Ordering::Greater => node.left = Self::delete_node(node.left.clone(), key), } - root.clone() + drop(node); + root } } ``` From 94f79ada2d4c234a7b8a67cddda169ef3c591ccf Mon Sep 17 00:00:00 2001 From: coffeelize <46947638+coffeelize@users.noreply.github.com> Date: Sun, 19 Mar 2023 11:13:54 +0800 Subject: [PATCH 003/154] =?UTF-8?q?Update=200001.=E4=B8=A4=E6=95=B0?= =?UTF-8?q?=E4=B9=8B=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 还有 --> 还要 --- problems/0001.两数之和.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md index a0acdcbe..a8f8bf17 100644 --- a/problems/0001.两数之和.md +++ b/problems/0001.两数之和.md @@ -41,7 +41,7 @@ 那么我们就应该想到使用哈希法了。 -因为本地，我们不仅要知道元素有没有遍历过，还有知道这个元素对应的下标，**需要使用 key value结构来存放，key来存元素，value来存下标，那么使用map正合适**。 +因为本地，我们不仅要知道元素有没有遍历过，还要知道这个元素对应的下标，**需要使用 key value结构来存放，key来存元素，value来存下标，那么使用map正合适**。 再来看一下使用数组和set来做哈希法的局限。 From 633cb549fb21c4f45a366a973dcc30768a97a03f Mon Sep 17 00:00:00 2001 From: Javie Deng Date: Mon, 20 Mar 2023 12:36:06 +0800 Subject: [PATCH 004/154] =?UTF-8?q?=E4=B8=BA0226.=E7=BF=BB=E8=BD=AC?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E6=B7=BB=E5=8A=A0=E4=BA=86C#?= =?UTF-8?q?=E8=AF=AD=E8=A8=80=E7=89=88=EF=BC=88=E5=8C=85=E5=90=AB=E9=80=92?= =?UTF-8?q?=E5=BD=92=E5=92=8C=E8=BF=AD=E4=BB=A3=EF=BC=89?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0226.翻转二叉树.md | 47 ++++++++++++++++++++++++++++++++ 1 file changed, 47 insertions(+) diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md index 16a5be57..611e971a 100644 --- a/problems/0226.翻转二叉树.md +++ b/problems/0226.翻转二叉树.md @@ -914,6 +914,53 @@ impl Solution { } ``` +### C# + +```csharp +//递归 +public class Solution { + public TreeNode InvertTree(TreeNode root) { + if (root == null) return root; + + swap(root); + InvertTree(root.left); + InvertTree(root.right); + return root; + } + + public void swap(TreeNode node) { + TreeNode temp = node.left; + node.left = node.right; + node.right = temp; + } +} +``` + +```csharp +//迭代 +public class Solution { + public TreeNode InvertTree(TreeNode root) { + if (root == null) return null; + Stack stack=new Stack(); + stack.Push(root); + while(stack.Count>0) + { + TreeNode node = stack.Pop(); + swap(node); + if(node.right!=null) stack.Push(node.right); + if(node.left!=null) stack.Push(node.left); + } + return root; + } + + public void swap(TreeNode node) { + TreeNode temp = node.left; + node.left = node.right; + node.right = temp; + } +} +``` +

From da2acd7dcf6c0a3fde44bb3801395eb7050f19f3 Mon Sep 17 00:00:00 2001 From: Javie Deng Date: Mon, 20 Mar 2023 13:05:00 +0800 Subject: [PATCH 005/154] =?UTF-8?q?"=E4=B8=BA0226.=E7=BF=BB=E8=BD=AC?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E6=B7=BB=E5=8A=A0=E4=BA=86C#?= =?UTF-8?q?=E8=AF=AD=E8=A8=80=E7=89=88=EF=BC=88=E5=8C=85=E5=90=AB=E9=80=92?= =?UTF-8?q?=E5=BD=92=E5=92=8C=E8=BF=AD=E4=BB=A3=EF=BC=89"?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0226.翻转二叉树.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md index 611e971a..c1f0f2c1 100644 --- a/problems/0226.翻转二叉树.md +++ b/problems/0226.翻转二叉树.md @@ -928,7 +928,7 @@ public class Solution { return root; } - public void swap(TreeNode node) { + public void swap(TreeNode node) { TreeNode temp = node.left; node.left = node.right; node.right = temp; @@ -953,7 +953,7 @@ public class Solution { return root; } - public void swap(TreeNode node) { + public void swap(TreeNode node) { TreeNode temp = node.left; node.left = node.right; node.right = temp; From f51f31290f69bf467bdb444d8c85a94b0c30616a Mon Sep 17 00:00:00 2001 From: fw_qaq Date: Thu, 23 Mar 2023 11:10:20 +0800 Subject: [PATCH 006/154] =?UTF-8?q?Update=200406.=E6=A0=B9=E6=8D=AE?= =?UTF-8?q?=E8=BA=AB=E9=AB=98=E9=87=8D=E5=BB=BA=E9=98=9F=E5=88=97.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0406.根据身高重建队列.md | 18 +++++++++--------- 1 file changed, 9 insertions(+), 9 deletions(-) diff --git a/problems/0406.根据身高重建队列.md b/problems/0406.根据身高重建队列.md index 48c498e1..2e6818db 100644 --- a/problems/0406.根据身高重建队列.md +++ b/problems/0406.根据身高重建队列.md @@ -291,19 +291,19 @@ var reconstructQueue = function(people) { ```Rust impl Solution { - pub fn reconstruct_queue(people: Vec>) -> Vec> { - let mut people = people; + pub fn reconstruct_queue(mut people: Vec>) -> Vec> { + let mut queue = vec![]; people.sort_by(|a, b| { - if a[0] == b[0] { return a[1].cmp(&b[1]); } + if a[0] == b[0] { + return a[1].cmp(&b[1]); + } b[0].cmp(&a[0]) }); - let mut que: Vec> = Vec::new(); - que.push(people[0].clone()); - for i in 1..people.len() { - let position = people[i][1]; - que.insert(position as usize, people[i].clone()); + queue.push(people[0].clone()); + for v in people.iter().skip(1) { + queue.insert(v[1] as usize, v.clone()); } - que + queue } } ``` From b934f47c83340b91b8a6291d96420978da0ec0e6 Mon Sep 17 00:00:00 2001 From: milu-tao <91822069+milu-tao@users.noreply.github.com> Date: Fri, 24 Mar 2023 22:14:39 +0800 Subject: [PATCH 007/154] =?UTF-8?q?Update=200647.=E5=9B=9E=E6=96=87?= =?UTF-8?q?=E5=AD=90=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 1. result 在遍历过程中 ++ 就行，没必要另外计算 2.题目已经表明：s字符串的长度规则是：1 <= s.length <= 1000，所以前面不需要进行判断字符串长度 --- problems/0647.回文子串.md | 33 ++++++++++++--------------------- 1 file changed, 12 insertions(+), 21 deletions(-) diff --git a/problems/0647.回文子串.md b/problems/0647.回文子串.md index 90e6da9f..339247f5 100644 --- a/problems/0647.回文子串.md +++ b/problems/0647.回文子串.md @@ -238,33 +238,24 @@ Java： ```java class Solution { public int countSubstrings(String s) { - int len, ans = 0; - if (s == null || (len = s.length()) < 1) return 0; - //dp[i][j]：s字符串下标i到下标j的字串是否是一个回文串，即s[i, j] + char[] chars = s.toCharArray(); + int len = chars.length; boolean[][] dp = new boolean[len][len]; - for (int j = 0; j < len; j++) { - for (int i = 0; i <= j; i++) { - //当两端字母一样时，才可以两端收缩进一步判断 - if (s.charAt(i) == s.charAt(j)) { - //i++，j--，即两端收缩之后i,j指针指向同一个字符或者i超过j了,必然是一个回文串 - if (j - i < 3) { + int result = 0; + for (int i = len - 1; i >= 0; i--) { + for (int j = i; j < len; j++) { + if (chars[i] == chars[j]) { + if (j - i <= 1) { // 情况一 和 情况二 + result++; + dp[i][j] = true; + } else if (dp[i + 1][j - 1]) { //情况三 + result++; dp[i][j] = true; - } else { - //否则通过收缩之后的字串判断 - dp[i][j] = dp[i + 1][j - 1]; } - } else {//两端字符不一样，不是回文串 - dp[i][j] = false; } } } - //遍历每一个字串，统计回文串个数 - for (int i = 0; i < len; i++) { - for (int j = 0; j < len; j++) { - if (dp[i][j]) ans++; - } - } - return ans; + return result; } } ``` From 5daf942cd584259762907c1e796ad535d5aa9ce6 Mon Sep 17 00:00:00 2001 From: Lozakaka <102352821+Lozakaka@users.noreply.github.com> Date: Wed, 29 Mar 2023 22:47:54 -0400 Subject: [PATCH 008/154] =?UTF-8?q?Update=200225.=E7=94=A8=E9=98=9F?= =?UTF-8?q?=E5=88=97=E5=AE=9E=E7=8E=B0=E6=A0=88.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 新增一個java solution用的是卡哥的邏輯 --- problems/0225.用队列实现栈.md | 37 +++++++++++++++++++++++++++++ 1 file changed, 37 insertions(+) diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md index 16b9e47c..66d807c1 100644 --- a/problems/0225.用队列实现栈.md +++ b/problems/0225.用队列实现栈.md @@ -324,6 +324,43 @@ class MyStack { } } +``` +优化，使用一个 Queue 实现，但用卡哥的逻辑实现 +``` +class MyStack { + Queue queue; + + public MyStack() { + queue = new LinkedList<>(); + } + + public void push(int x) { + queue.add(x); + } + + public int pop() { + rePosition(); + return queue.poll(); + } + + public int top() { + rePosition(); + int result = queue.poll(); + queue.add(result); + return result; + } + + public boolean empty() { + return queue.isEmpty(); + } + + public void rePosition(){ + int size = queue.size(); + size--; + while(size-->0) + queue.add(queue.poll()); + } +} ``` Python： From 872402e5a4e2e227e889fd3103786c503f864abd Mon Sep 17 00:00:00 2001 From: fw_qaq Date: Thu, 13 Apr 2023 18:05:44 +0800 Subject: [PATCH 009/154] =?UTF-8?q?Update=200063.=E4=B8=8D=E5=90=8C?= =?UTF-8?q?=E8=B7=AF=E5=BE=84II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0063.不同路径II.md | 28 ++++++++++++++++++++++++++++ 1 file changed, 28 insertions(+) diff --git a/problems/0063.不同路径II.md b/problems/0063.不同路径II.md index 85130ab4..8ab90e6d 100644 --- a/problems/0063.不同路径II.md +++ b/problems/0063.不同路径II.md @@ -493,6 +493,34 @@ impl Solution { } ``` +空间优化： + +```rust +impl Solution { + pub fn unique_paths_with_obstacles(obstacle_grid: Vec>) -> i32 { + let mut dp = vec![0; obstacle_grid[0].len()]; + for (i, &v) in obstacle_grid[0].iter().enumerate() { + if v == 0 { + dp[i] = 1; + } else { + break; + } + } + for rows in obstacle_grid.iter().skip(1) { + for j in 0..rows.len() { + if rows[j] == 1 { + dp[j] = 0; + continue; + } else if j != 0 { + dp[j] += dp[j - 1]; + } + } + } + dp.pop().unwrap() + } +} +``` + ### C ```c From ced650c7bb9d43447e8445fff31469455440057e Mon Sep 17 00:00:00 2001 From: fw_qaq Date: Thu, 13 Apr 2023 18:06:59 +0800 Subject: [PATCH 010/154] =?UTF-8?q?Update=20problems/0063.=E4=B8=8D?= =?UTF-8?q?=E5=90=8C=E8=B7=AF=E5=BE=84II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0063.不同路径II.md | 1 - 1 file changed, 1 deletion(-) diff --git a/problems/0063.不同路径II.md b/problems/0063.不同路径II.md index 8ab90e6d..8ff8c33f 100644 --- a/problems/0063.不同路径II.md +++ b/problems/0063.不同路径II.md @@ -510,7 +510,6 @@ impl Solution { for j in 0..rows.len() { if rows[j] == 1 { dp[j] = 0; - continue; } else if j != 0 { dp[j] += dp[j - 1]; } From bd22ad9257b90dbeb2db60f5cdd6c591f6288e37 Mon Sep 17 00:00:00 2001 From: fw_qaq Date: Thu, 13 Apr 2023 18:51:26 +0800 Subject: [PATCH 011/154] =?UTF-8?q?Update=200343.=E6=95=B4=E6=95=B0?= =?UTF-8?q?=E6=8B=86=E5=88=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0343.整数拆分.md | 44 ++++++++++++++++++----------------- 1 file changed, 23 insertions(+), 21 deletions(-) diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md index c2fbbdde..4df6c41f 100644 --- a/problems/0343.整数拆分.md +++ b/problems/0343.整数拆分.md @@ -319,6 +319,29 @@ pub fn integer_break(n: i32) -> i32 { } ``` +贪心： + +```rust +impl Solution { + pub fn integer_break(mut n: i32) -> i32 { + match n { + 2 => 1, + 3 => 2, + 4 => 4, + 5.. => { + let mut res = 1; + while n > 4 { + res *= 3; + n -= 3; + } + res * n + } + _ => panic!("Error"), + } + } +} +``` + ### TypeScript ```typescript @@ -344,27 +367,6 @@ function integerBreak(n: number): number { }; ``` -### Rust - -```Rust -impl Solution { - fn max(a: i32, b: i32) -> i32{ - if a > b { a } else { b } - } - pub fn integer_break(n: i32) -> i32 { - let n = n as usize; - let mut dp = vec![0; n + 1]; - dp[2] = 1; - for i in 3..=n { - for j in 1..i - 1 { - dp[i] = Self::max(dp[i], Self::max(((i - j) * j) as i32, dp[i - j] * j as i32)); - } - } - dp[n] - } -} -``` - ### C ```c From 7111943ac00f9cea2c4899f0211b2d0fcadeef62 Mon Sep 17 00:00:00 2001 From: BanTanger <88583317+BanTanger@users.noreply.github.com> Date: Thu, 13 Apr 2023 22:18:21 +0800 Subject: [PATCH 012/154] =?UTF-8?q?Update=200347.=E5=89=8DK=E4=B8=AA?= =?UTF-8?q?=E9=AB=98=E9=A2=91=E5=85=83=E7=B4=A0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 之前的 java 代码调用的 api 过于难记忆，不利于白板书写，于是将优先级队列存储数据结构从 map.setEntry 改为 int[] 数组，方便大家记忆理解书写 --- problems/0347.前K个高频元素.md | 28 +++++++++++++++++++++++++++- 1 file changed, 27 insertions(+), 1 deletion(-) diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md index 0d268d9b..6c8b51b1 100644 --- a/problems/0347.前K个高频元素.md +++ b/problems/0347.前K个高频元素.md @@ -188,7 +188,33 @@ class Solution { } } ``` - +简化版代码： +```java +class Solution { + public int[] topKFrequent(int[] nums, int k) { + // 优先级队列，为了避免复杂 api 操作，pq 存储数组 + // lambda 表达式设置优先级队列从大到小存储 o1 - o2 为从大到小，o2 - o1 反之 + PriorityQueue pq = new PriorityQueue<>((o1, o2) -> o1[1] - o2[1]); + int[] res = new int[k]; // 答案数组为 k 个元素 + Map map = new HashMap<>(); // 记录元素出现次数 + for(int num : nums) map.put(num, map.getOrDefault(num, 0) + 1); + for(var x : map.entrySet()) { // entrySet 获取 k-v Set 集合 + // 将 kv 转化成数组 + int[] tmp = new int[2]; + tmp[0] = x.getKey(); + tmp[1] = x.getValue(); + pq.offer(tmp); + if(pq.size() > k) { + pq.poll(); + } + } + for(int i = 0; i < k; i ++) { + res[i] = pq.poll()[0]; // 获取优先队列里的元素 + } + return res; + } +} +``` Python： ```python From 9aaf02cc20156d385853f0d3f8c9eebc983bc9d7 Mon Sep 17 00:00:00 2001 From: sharky <1821984081@qq.com> Date: Tue, 18 Apr 2023 15:15:28 +0800 Subject: [PATCH 013/154] =?UTF-8?q?1005=E3=80=81K=E6=AC=A1=E5=8F=96?= =?UTF-8?q?=E5=8F=8D=E5=90=8E=E6=9C=80=E5=A4=A7=E5=8C=96=E7=9A=84=E6=95=B0?= =?UTF-8?q?=E7=BB=84=E5=92=8C=EF=BC=8C=E6=9A=B4=E5=8A=9B=E8=A7=A3=E6=B3=95?= =?UTF-8?q?=EF=BC=8Cjava?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../1005.K次取反后最大化的数组和.md | 18 +++++++++++++++++- 1 file changed, 17 insertions(+), 1 deletion(-) diff --git a/problems/1005.K次取反后最大化的数组和.md b/problems/1005.K次取反后最大化的数组和.md index cdf42511..18b07b89 100644 --- a/problems/1005.K次取反后最大化的数组和.md +++ b/problems/1005.K次取反后最大化的数组和.md @@ -145,7 +145,23 @@ class Solution { } } ``` - +```java +//暴力解法 +class Solution { + public int largestSumAfterKNegations(int[] nums, int k) { + int count = 0; + //循环k次 + for( int i = 0; i < k; i++ ){ + Arrays.sort( nums ); + nums[0] = -nums[0];//把最小值换成其相反数 + } + for ( int num : nums ){ + count += num;//累加 + } + return count; + } +} +``` ### Python ```python class Solution: From b7ea55c93b68391ecd87c39c06e2811f0ed6b98f Mon Sep 17 00:00:00 2001 From: Winson Huang Date: Tue, 18 Apr 2023 21:07:54 +0800 Subject: [PATCH 014/154] =?UTF-8?q?Update=200454.=E5=9B=9B=E6=95=B0?= =?UTF-8?q?=E7=9B=B8=E5=8A=A0II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改 Java 语言版本代码，让 HashMap 相关操作更简洁 --- problems/0454.四数相加II.md | 13 +++---------- 1 file changed, 3 insertions(+), 10 deletions(-) diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index 411b60e8..abfc7c23 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -102,21 +102,14 @@ class Solution { //统计两个数组中的元素之和，同时统计出现的次数，放入map for (int i : nums1) { for (int j : nums2) { - int tmp = map.getOrDefault(i + j, 0); - if (tmp == 0) { - map.put(i + j, 1); - } else { - map.replace(i + j, tmp + 1); - } + int sum = i + j; + map.put(sum, map.getOrDefault(sum, 0) + 1); } } //统计剩余的两个元素的和，在map中找是否存在相加为0的情况，同时记录次数 for (int i : nums3) { for (int j : nums4) { - int tmp = map.getOrDefault(0 - i - j, 0); - if (tmp != 0) { - res += tmp; - } + res += map.getOrDefault(0 - i - j, 0); } } return res; From 747563607292b4fc84452c4f45c2d0504531a171 Mon Sep 17 00:00:00 2001 From: Winson Huang Date: Tue, 18 Apr 2023 21:19:42 +0800 Subject: [PATCH 015/154] =?UTF-8?q?Update=200383.=E8=B5=8E=E9=87=91?= =?UTF-8?q?=E4=BF=A1.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 为 Java 语言版本的代码添加长度判断 --- problems/0383.赎金信.md | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index d9a184b6..a3f87d4a 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -117,6 +117,10 @@ Java： ```Java class Solution { public boolean canConstruct(String ransomNote, String magazine) { + // shortcut + if (ransomNote.length() > magazine.length()) { + return false; + } // 定义一个哈希映射数组 int[] record = new int[26]; From 3f2a816f3096c6ccccba41c81b3975ca80f6d081 Mon Sep 17 00:00:00 2001 From: Lozakaka <102352821+Lozakaka@users.noreply.github.com> Date: Tue, 18 Apr 2023 18:23:24 -0400 Subject: [PATCH 016/154] =?UTF-8?q?Update=200112.=E8=B7=AF=E5=BE=84?= =?UTF-8?q?=E6=80=BB=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 新增java 統一迭代法 --- problems/0112.路径总和.md | 36 +++++++++++++++++++++++++++++++++++ 1 file changed, 36 insertions(+) diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index 5958de93..b1457887 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -385,6 +385,42 @@ class solution { } } ``` +```Java 統一迭代法 + public boolean hasPathSum(TreeNode root, int targetSum) { + Stack treeNodeStack = new Stack<>(); + Stack sumStack = new Stack<>(); + + if(root == null) + return false; + treeNodeStack.add(root); + sumStack.add(root.val); + + while(!treeNodeStack.isEmpty()){ + TreeNode curr = treeNodeStack.peek(); + int tempsum = sumStack.pop(); + if(curr != null){ + treeNodeStack.pop(); + treeNodeStack.add(curr); + treeNodeStack.add(null); + sumStack.add(tempsum); + if(curr.right != null){ + treeNodeStack.add(curr.right); + sumStack.add(tempsum + curr.right.val); + } + if(curr.left != null){ + treeNodeStack.add(curr.left); + sumStack.add(tempsum + curr.left.val); + } + }else{ + treeNodeStack.pop(); + TreeNode temp = treeNodeStack.pop(); + if(temp.left == null && temp.right == null && tempsum == targetSum) + return true; + } + } + return false; + } +``` ### 0113.路径总和-ii From 0cfe55180439c77995abefaa02c5b61108203f3a Mon Sep 17 00:00:00 2001 From: asxy <17375702582@163.com> Date: Thu, 20 Apr 2023 11:14:43 +0800 Subject: [PATCH 017/154] =?UTF-8?q?Update=200647.=E5=9B=9E=E6=96=87?= =?UTF-8?q?=E5=AD=90=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 回文子串添加java动态规划简单版本的代码 --- problems/0647.回文子串.md | 21 +++++++++++++++++++++ 1 file changed, 21 insertions(+) diff --git a/problems/0647.回文子串.md b/problems/0647.回文子串.md index 90e6da9f..521c8c26 100644 --- a/problems/0647.回文子串.md +++ b/problems/0647.回文子串.md @@ -267,6 +267,27 @@ class Solution { return ans; } } + +``` + +动态规划：简洁版 +```java +class Solution { + public int countSubstrings(String s) { + boolean[][] dp = new boolean[s.length()][s.length()]; + + int res = 0; + for (int i = s.length() - 1; i >= 0; i--) { + for (int j = i; j < s.length(); j++) { + if (s.charAt(i) == s.charAt(j) && (j - i <= 1 || dp[i + 1][j - 1])) { + res++; + dp[i][j] = true; + } + } + } + return res; + } +} ``` 中心扩散法： From fca305039005936bc9678c01d7573d6a2f8552ba Mon Sep 17 00:00:00 2001 From: asxy Date: Thu, 20 Apr 2023 11:47:39 +0800 Subject: [PATCH 018/154] =?UTF-8?q?Update=200084.=E6=9F=B1=E7=8A=B6?= =?UTF-8?q?=E5=9B=BE=E4=B8=AD=E6=9C=80=E5=A4=A7=E7=9A=84=E7=9F=A9=E5=BD=A2?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加单调栈精简java代码 --- problems/0084.柱状图中最大的矩形.md | 27 ++++++++++++++++++++ 1 file changed, 27 insertions(+) diff --git a/problems/0084.柱状图中最大的矩形.md b/problems/0084.柱状图中最大的矩形.md index eb064143..f9a83508 100644 --- a/problems/0084.柱状图中最大的矩形.md +++ b/problems/0084.柱状图中最大的矩形.md @@ -307,6 +307,33 @@ class Solution { } } ``` +单调栈精简 +```java +class Solution { + public int largestRectangleArea(int[] heights) { + int[] newHeight = new int[heights.length + 2]; + System.arraycopy(heights, 0, newHeight, 1, heights.length); + newHeight[heights.length+1] = 0; + newHeight[0] = 0; + + Stack stack = new Stack<>(); + stack.push(0); + + int res = 0; + for (int i = 1; i < newHeight.length; i++) { + while (newHeight[i] < newHeight[stack.peek()]) { + int mid = stack.pop(); + int w = i - stack.peek() - 1; + int h = newHeight[mid]; + res = Math.max(res, w * h); + } + stack.push(i); + + } + return res; + } +} +``` Python3: From f7bdc58b2802e8b81483a5c2f8023499a3d525c4 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 20 Apr 2023 01:18:03 -0500 Subject: [PATCH 019/154] =?UTF-8?q?Update=200707.=E8=AE=BE=E8=AE=A1?= =?UTF-8?q?=E9=93=BE=E8=A1=A8.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改 python 代码，dummy node --- problems/0707.设计链表.md | 71 +++++++++++++++++++---------------- 1 file changed, 38 insertions(+), 33 deletions(-) diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index 5c78a12a..c72b7327 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -486,15 +486,10 @@ class MyLinkedList { Python： ```python # 单链表 -class Node(object): - def __init__(self, x=0): - self.val = x - self.next = None - -class MyLinkedList(object): +class MyLinkedList1: def __init__(self): - self.head = Node() + self.dummy_head = Node()# 添加虚拟头指针，便于操作 self.size = 0 # 设置一个链表长度的属性，便于后续操作，注意每次增和删的时候都要更新 def get(self, index): @@ -504,7 +499,7 @@ class MyLinkedList(object): """ if index < 0 or index >= self.size: return -1 - cur = self.head.next + cur = self.dummy_head.next while(index): cur = cur.next index -= 1 @@ -516,8 +511,8 @@ class MyLinkedList(object): :rtype: None """ new_node = Node(val) - new_node.next = self.head.next - self.head.next = new_node + new_node.next = self.dummy_head.next + self.dummy_head.next = new_node self.size += 1 def addAtTail(self, val): @@ -526,7 +521,7 @@ class MyLinkedList(object): :rtype: None """ new_node = Node(val) - cur = self.head + cur = self.dummy_head while(cur.next): cur = cur.next cur.next = new_node @@ -548,12 +543,12 @@ class MyLinkedList(object): return node = Node(val) - pre = self.head + cur = self.dummy_head while(index): - pre = pre.next + cur = cur.next index -= 1 - node.next = pre.next - pre.next = node + node.next = cur.next + cur.next = node self.size += 1 def deleteAtIndex(self, index): @@ -563,7 +558,7 @@ class MyLinkedList(object): """ if index < 0 or index >= self.size: return - pre = self.head + pre = self.dummy_head while(index): pre = pre.next index -= 1 @@ -574,11 +569,10 @@ class MyLinkedList(object): # 相对于单链表, Node新增了prev属性 class Node: - def __init__(self, val): + def __init__(self, val=0, next = None, prev = None): self.val = val - self.prev = None - self.next = None - + self.next = next + self.prev = prev class MyLinkedList: @@ -601,6 +595,20 @@ class MyLinkedList: node = node.next return node + + def _update(self, prev: Node, next: Node, val: int) -> None: + """ + 更新节点 + :param prev: 相对于更新的前一个节点 + :param next: 相对于更新的后一个节点 + :param val: 要添加的节点值 + """ + # 计数累加 + self._count += 1 + node = Node(val) + prev.next, next.prev = node, node + node.prev, node.next = prev, next + def get(self, index: int) -> int: """ Get the value of the index-th node in the linked list. If the index is invalid, return -1. @@ -634,19 +642,6 @@ class MyLinkedList: node = self._get_node(index) self._update(node.prev, node, val) - def _update(self, prev: Node, next: Node, val: int) -> None: - """ - 更新节点 - :param prev: 相对于更新的前一个节点 - :param next: 相对于更新的后一个节点 - :param val: 要添加的节点值 - """ - # 计数累加 - self._count += 1 - node = Node(val) - prev.next, next.prev = node, node - node.prev, node.next = prev, next - def deleteAtIndex(self, index: int) -> None: """ Delete the index-th node in the linked list, if the index is valid. @@ -656,6 +651,16 @@ class MyLinkedList: # 计数-1 self._count -= 1 node.prev.next, node.next.prev = node.next, node.prev + + + +# Your MyLinkedList object will be instantiated and called as such: +# obj = MyLinkedList() +# param_1 = obj.get(index) +# obj.addAtHead(val) +# obj.addAtTail(val) +# obj.addAtIndex(index,val) +# obj.deleteAtIndex(index) ``` Go： From ad2acdb14ac6d633a0ad6e3b8f92d3e1f6c32881 Mon Sep 17 00:00:00 2001 From: Lozakaka <102352821+Lozakaka@users.noreply.github.com> Date: Thu, 20 Apr 2023 23:44:00 -0400 Subject: [PATCH 020/154] =?UTF-8?q?Update=200112.=E8=B7=AF=E5=BE=84?= =?UTF-8?q?=E6=80=BB=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit add java iteration method for leetcode 113 (DFS统一迭代法) --- problems/0112.路径总和.md | 46 +++++++++++++++++++++++++++++++++++ 1 file changed, 46 insertions(+) diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index 5958de93..d8ea0a18 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -446,6 +446,52 @@ class Solution { } } ``` +```java +// 解法3 DFS统一迭代法 +class Solution { + public List> pathSum(TreeNode root, int targetSum) { + List> result = new ArrayList<>(); + Stack nodeStack = new Stack<>(); + Stack sumStack = new Stack<>(); + Stack> pathStack = new Stack<>(); + if(root == null) + return result; + nodeStack.add(root); + sumStack.add(root.val); + pathStack.add(new ArrayList<>()); + + while(!nodeStack.isEmpty()){ + TreeNode currNode = nodeStack.peek(); + int currSum = sumStack.pop(); + ArrayList currPath = pathStack.pop(); + if(currNode != null){ + nodeStack.pop(); + nodeStack.add(currNode); + nodeStack.add(null); + sumStack.add(currSum); + currPath.add(currNode.val); + pathStack.add(new ArrayList(currPath)); + if(currNode.right != null){ + nodeStack.add(currNode.right); + sumStack.add(currSum + currNode.right.val); + pathStack.add(new ArrayList(currPath)); + } + if(currNode.left != null){ + nodeStack.add(currNode.left); + sumStack.add(currSum + currNode.left.val); + pathStack.add(new ArrayList(currPath)); + } + }else{ + nodeStack.pop(); + TreeNode temp = nodeStack.pop(); + if(temp.left == null && temp.right == null && currSum == targetSum) + result.add(new ArrayList(currPath)); + } + } + return result; + } +} +``` ## python From 252e330a6b20fa13e81e4412c91f40547e0ba1b1 Mon Sep 17 00:00:00 2001 From: HOUSHENGREN <48871516+HOUSHENGREN@users.noreply.github.com> Date: Sun, 23 Apr 2023 14:36:27 +0800 Subject: [PATCH 021/154] =?UTF-8?q?Update=20=E5=88=B7=E5=8A=9B=E6=89=A3?= =?UTF-8?q?=E7=94=A8=E4=B8=8D=E7=94=A8=E5=BA=93=E5=87=BD=E6=95=B0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit fix：文字错误 --- problems/前序/刷力扣用不用库函数.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/前序/刷力扣用不用库函数.md b/problems/前序/刷力扣用不用库函数.md index ae0940bf..04fce856 100644 --- a/problems/前序/刷力扣用不用库函数.md +++ b/problems/前序/刷力扣用不用库函数.md @@ -24,5 +24,5 @@ 例如for循环里套一个字符串的insert，erase之类的操作，你说时间复杂度是多少呢，很明显是O(n^2)的时间复杂度了。 -在刷题的时候本着我说的标准来使用库函数，详细对大家回有所帮助！ +在刷题的时候本着我说的标准来使用库函数，相信对大家回有所帮助！ From b8f6df60c432fd85d9417e0ce138a90b4603527e Mon Sep 17 00:00:00 2001 From: Lozakaka <102352821+Lozakaka@users.noreply.github.com> Date: Sun, 23 Apr 2023 02:40:34 -0400 Subject: [PATCH 022/154] =?UTF-8?q?Update=200106.=E4=BB=8E=E4=B8=AD?= =?UTF-8?q?=E5=BA=8F=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86=E5=BA=8F?= =?UTF-8?q?=E5=88=97=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 新增一java寫法 和卡哥的思路一樣的 --- ...序与后序遍历序列构造二叉树.md | 35 +++++++++++++++++++ 1 file changed, 35 insertions(+) diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index adb374f9..8fc973e0 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -622,7 +622,42 @@ class Solution { } } ``` +```java +class Solution { + public TreeNode buildTree(int[] inorder, int[] postorder) { + if(postorder.length == 0 || inorder.length == 0) + return null; + return buildHelper(inorder, 0, inorder.length, postorder, 0, postorder.length); + + } + private TreeNode buildHelper(int[] inorder, int inorderStart, int inorderEnd, int[] postorder, int postorderStart, int postorderEnd){ + if(postorderStart == postorderEnd) + return null; + int rootVal = postorder[postorderEnd - 1]; + TreeNode root = new TreeNode(rootVal); + int middleIndex; + for (middleIndex = inorderStart; middleIndex < inorderEnd; middleIndex++){ + if(inorder[middleIndex] == rootVal) + break; + } + int leftInorderStart = inorderStart; + int leftInorderEnd = middleIndex; + int rightInorderStart = middleIndex + 1; + int rightInorderEnd = inorderEnd; + + + int leftPostorderStart = postorderStart; + int leftPostorderEnd = postorderStart + (middleIndex - inorderStart); + int rightPostorderStart = leftPostorderEnd; + int rightPostorderEnd = postorderEnd - 1; + root.left = buildHelper(inorder, leftInorderStart, leftInorderEnd, postorder, leftPostorderStart, leftPostorderEnd); + root.right = buildHelper(inorder, rightInorderStart, rightInorderEnd, postorder, rightPostorderStart, rightPostorderEnd); + + return root; + } +} +``` 105.从前序与中序遍历序列构造二叉树 ```java From 9558af957bd2077773a11fb4ebae7e432838c08a Mon Sep 17 00:00:00 2001 From: Charlie <1753524606@qq.com> Date: Sun, 23 Apr 2023 14:55:28 +0800 Subject: [PATCH 023/154] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200028.=E5=AE=9E?= =?UTF-8?q?=E7=8E=B0strStr=20Rust=E7=89=88=E6=9C=AC=E5=A4=9A=E4=BD=99?= =?UTF-8?q?=E4=BB=A3=E7=A0=81=E5=88=A0=E9=99=A4?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0028.实现strStr.md | 4 ---- 1 file changed, 4 deletions(-) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index 263c1689..2a81db88 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -1314,7 +1314,6 @@ impl Solution { pub fn str_str(haystack: String, needle: String) -> i32 { let (haystack_len, needle_len) = (haystack.len(), needle.len()); - if haystack_len == 0 { return 0; } if haystack_len < needle_len { return -1;} let (haystack, needle) = (haystack.chars().collect::>(), needle.chars().collect::>()); let mut next: Vec = vec![0; haystack_len]; @@ -1355,9 +1354,6 @@ impl Solution { next } pub fn str_str(haystack: String, needle: String) -> i32 { - if needle.is_empty() { - return 0; - } if haystack.len() < needle.len() { return -1; } From f17c74ceec294a90f864a1a9a3836a03b4a752bb Mon Sep 17 00:00:00 2001 From: HOUSHENGREN <48871516+HOUSHENGREN@users.noreply.github.com> Date: Sun, 23 Apr 2023 15:14:25 +0800 Subject: [PATCH 024/154] =?UTF-8?q?Update=20=E5=88=B7=E5=8A=9B=E6=89=A3?= =?UTF-8?q?=E7=94=A8=E4=B8=8D=E7=94=A8=E5=BA=93=E5=87=BD=E6=95=B0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit feat：修改文案 --- problems/前序/刷力扣用不用库函数.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/前序/刷力扣用不用库函数.md b/problems/前序/刷力扣用不用库函数.md index 04fce856..7d0e3475 100644 --- a/problems/前序/刷力扣用不用库函数.md +++ b/problems/前序/刷力扣用不用库函数.md @@ -24,5 +24,5 @@ 例如for循环里套一个字符串的insert，erase之类的操作，你说时间复杂度是多少呢，很明显是O(n^2)的时间复杂度了。 -在刷题的时候本着我说的标准来使用库函数，相信对大家回有所帮助！ +在刷题的时候本着我说的标准来使用库函数，相信对大家会有所帮助！ From ad871dae77cfed61609d3bffd14e2fd994ae7d1d Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E5=93=88=E5=93=88=E5=93=88?= <76643786+Projecthappy@users.noreply.github.com> Date: Sun, 23 Apr 2023 18:22:29 +0800 Subject: [PATCH 025/154] =?UTF-8?q?Update=200225.=E7=94=A8=E9=98=9F?= =?UTF-8?q?=E5=88=97=E5=AE=9E=E7=8E=B0=E6=A0=88.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 为使用两个 Queue 实现添加新方法，方法将q1作为主要的队列，其元素排列顺序和出栈顺序相同，q2仅作为临时放置，push方法中在加入元素时先将q1中的元素依次出栈压入q2，然后将新加入的元素压入q1，再将q2中的元素依次出栈压入q1，其他方法可直接使用Queue中已有的方法。 --- problems/0225.用队列实现栈.md | 38 ++++++++++++++++++++++++++++- 1 file changed, 37 insertions(+), 1 deletion(-) diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md index bad2faec..29ef0933 100644 --- a/problems/0225.用队列实现栈.md +++ b/problems/0225.用队列实现栈.md @@ -166,7 +166,7 @@ public: Java： -使用两个 Queue 实现 +使用两个 Queue 实现方法1 ```java class MyStack { @@ -208,6 +208,42 @@ class MyStack { } ``` +使用两个 Queue 实现方法2 +```java +class MyStack { + //q1作为主要的队列，其元素排列顺序和出栈顺序相同 + Queue q1 = new ArrayDeque<>(); + //q2仅作为临时放置 + Queue q2 = new ArrayDeque<>(); + + public MyStack() { + + } + //在加入元素时先将q1中的元素依次出栈压入q2，然后将新加入的元素压入q1，再将q2中的元素依次出栈压入q1 + public void push(int x) { + while (q1.size() > 0) { + q2.add(q1.poll()); + } + q1.add(x); + while (q2.size() > 0) { + q1.add(q2.poll()); + } + } + + public int pop() { + return q1.poll(); + } + + public int top() { + return q1.peek(); + } + + public boolean empty() { + return q1.isEmpty(); + } +} +``` + 使用两个 Deque 实现 ```java class MyStack { From ce73cef055f17879a9ef8f58f8cabffd48ed7edb Mon Sep 17 00:00:00 2001 From: jimowo <1252480844@qq.com> Date: Mon, 24 Apr 2023 16:16:16 +0800 Subject: [PATCH 026/154] =?UTF-8?q?=E7=AE=97=E6=B3=95=E7=A9=BA=E9=97=B4?= =?UTF-8?q?=E4=BC=98=E5=8C=96?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0198.打家劫舍.md | 24 ++++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/0198.打家劫舍.md b/problems/0198.打家劫舍.md index c25f3b86..6e7f5ab6 100644 --- a/problems/0198.打家劫舍.md +++ b/problems/0198.打家劫舍.md @@ -136,6 +136,29 @@ class Solution { return dp[nums.length - 1]; } } + +// 空间优化 dp数组只存与计算相关的两次数据 +class Solution { + public int rob(int[] nums) { + if (nums.length == 1) { + return nums[0]; + } + // 初始化dp数组 + // 优化空间 dp数组只用2格空间 只记录与当前计算相关的前两个结果 + int[] dp = new int[2]; + dp[0] = nums[0]; + dp[1] = nums[0] > nums[1] ? nums[0] : nums[1]; + int res = 0; + // 遍历 + for (int i = 2; i < nums.length; i++) { + res = (dp[0] + nums[i]) > dp[1] ? (dp[0] + nums[i]) : dp[1]; + dp[0] = dp[1]; + dp[1] = res; + } + // 输出结果 + return dp[1]; + } +} ``` Python： @@ -220,3 +243,4 @@ function rob(nums: number[]): number { + From bb8e7312b64b13f98ca6c6ae88260a4f57375acd Mon Sep 17 00:00:00 2001 From: Lozakaka <102352821+Lozakaka@users.noreply.github.com> Date: Tue, 25 Apr 2023 03:25:45 -0400 Subject: [PATCH 027/154] =?UTF-8?q?Update=200098.=E9=AA=8C=E8=AF=81?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 新增java 統一迭代法的寫法 有通過. AC --- problems/0098.验证二叉搜索树.md | 30 ++++++++++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/problems/0098.验证二叉搜索树.md b/problems/0098.验证二叉搜索树.md index 95afe680..ccb63305 100644 --- a/problems/0098.验证二叉搜索树.md +++ b/problems/0098.验证二叉搜索树.md @@ -259,6 +259,36 @@ public: ## Java +```Java +//使用統一迭代法 +class Solution { + public boolean isValidBST(TreeNode root) { + Stack stack = new Stack<>(); + TreeNode pre = null; + if(root != null) + stack.add(root); + while(!stack.isEmpty()){ + TreeNode curr = stack.peek(); + if(curr != null){ + stack.pop(); + if(curr.right != null) + stack.add(curr.right); + stack.add(curr); + stack.add(null); + if(curr.left != null) + stack.add(curr.left); + }else{ + stack.pop(); + TreeNode temp = stack.pop(); + if(pre != null && pre.val >= temp.val) + return false; + pre = temp; + } + } + return true; + } +} +``` ```Java class Solution { // 递归 From 75c7f940d112893dd6a1f89f844b710a87b202e1 Mon Sep 17 00:00:00 2001 From: Lozakaka <102352821+Lozakaka@users.noreply.github.com> Date: Tue, 25 Apr 2023 04:03:17 -0400 Subject: [PATCH 028/154] =?UTF-8?q?=E6=96=B0=E5=A2=9Ejava=20=E7=B5=B1?= =?UTF-8?q?=E4=B8=80=E8=BF=AD=E4=BB=A3=E6=B3=95-=E4=B8=AD=E5=BA=8F?= =?UTF-8?q?=E9=81=8D=E5=8E=86?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 新增java 統一迭代法-中序遍历 --- .../0530.二叉搜索树的最小绝对差.md | 33 +++++++++++++++++++ 1 file changed, 33 insertions(+) diff --git a/problems/0530.二叉搜索树的最小绝对差.md b/problems/0530.二叉搜索树的最小绝对差.md index fa1430de..cd24c6fa 100644 --- a/problems/0530.二叉搜索树的最小绝对差.md +++ b/problems/0530.二叉搜索树的最小绝对差.md @@ -174,6 +174,39 @@ class Solution { } } ``` +統一迭代法-中序遍历 +```Java +class Solution { + public int getMinimumDifference(TreeNode root) { + Stack stack = new Stack<>(); + TreeNode pre = null; + int result = Integer.MAX_VALUE; + + if(root != null) + stack.add(root); + while(!stack.isEmpty()){ + TreeNode curr = stack.peek(); + if(curr != null){ + stack.pop(); + if(curr.right != null) + stack.add(curr.right); + stack.add(curr); + stack.add(null); + if(curr.left != null) + stack.add(curr.left); + }else{ + stack.pop(); + TreeNode temp = stack.pop(); + if(pre != null) + result = Math.min(result, temp.val - pre.val); + pre = temp; + } + } + return result; + } +} +``` + 迭代法-中序遍历 ```java From c5c7dfb7fe0c0e7aa5c2115bed085d417d0726c0 Mon Sep 17 00:00:00 2001 From: zbb Date: Tue, 25 Apr 2023 19:45:48 +0800 Subject: [PATCH 029/154] =?UTF-8?q?=E4=BF=AE=E5=A4=8D=200101.=E5=AF=B9?= =?UTF-8?q?=E7=A7=B0=E4=BA=8C=E5=8F=89=E6=A0=91.md=20=E9=87=8C=E7=9A=84?= =?UTF-8?q?=E9=94=99=E5=88=AB=E5=AD=97?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0101.对称二叉树.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md index b75e9ff2..1594196b 100644 --- a/problems/0101.对称二叉树.md +++ b/problems/0101.对称二叉树.md @@ -88,7 +88,7 @@ else if (left->val != right->val) return false; // 注意这里我没有 * 比较二叉树外侧是否对称：传入的是左节点的左孩子，右节点的右孩子。 -* 比较内测是否对称，传入左节点的右孩子，右节点的左孩子。 +* 比较内侧是否对称，传入左节点的右孩子，右节点的左孩子。 * 如果左右都对称就返回true ，有一侧不对称就返回false 。 代码如下： @@ -157,7 +157,7 @@ public: **这个代码就很简洁了，但隐藏了很多逻辑，条理不清晰，而且递归三部曲，在这里完全体现不出来。** -**所以建议大家做题的时候，一定要想清楚逻辑，每一步做什么。把道题目所有情况想到位，相应的代码写出来之后，再去追求简洁代码的效果。** +**所以建议大家做题的时候，一定要想清楚逻辑，每一步做什么。把题目所有情况想到位，相应的代码写出来之后，再去追求简洁代码的效果。** ## 迭代法 From 7851e51c7345590977d3881424354d6be0dd8eae Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 15:35:27 +0800 Subject: [PATCH 030/154] =?UTF-8?q?update=20=200077.=E7=BB=84=E5=90=88?= =?UTF-8?q?=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82=E5=BA=A6=E5=88=86?= =?UTF-8?q?=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0077.组合.md | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/problems/0077.组合.md b/problems/0077.组合.md index 9c6d481d..3f222a17 100644 --- a/problems/0077.组合.md +++ b/problems/0077.组合.md @@ -218,6 +218,10 @@ public: } }; ``` +* 时间复杂度: O(n * 2^n) +* 空间复杂度: O(n) + + 还记得我们在[关于回溯算法，你该了解这些！](https://programmercarl.com/回溯算法理论基础.html)中给出的回溯法模板么？ From e533cb9cd4a5048a933055d5dd02ad7dbf8a1d20 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 15:37:21 +0800 Subject: [PATCH 031/154] =?UTF-8?q?update=20=20=200077.=E7=BB=84=E5=90=88?= =?UTF-8?q?=E4=BC=98=E5=8C=96=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82?= =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0077.组合优化.md | 4 ++++ 1 file changed, 4 insertions(+) diff --git a/problems/0077.组合优化.md b/problems/0077.组合优化.md index 9736549c..0c816bc1 100644 --- a/problems/0077.组合优化.md +++ b/problems/0077.组合优化.md @@ -130,6 +130,10 @@ public: } }; ``` +* 时间复杂度: O(n * 2^n) +* 空间复杂度: O(n) + + # 总结 From 9a29e5c3be75185c6013aaa0cb252374fb80598d Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 15:41:40 +0800 Subject: [PATCH 032/154] =?UTF-8?q?update=20=20=200216.=E7=BB=84=E5=90=88?= =?UTF-8?q?=E6=80=BB=E5=92=8CIII=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D?= =?UTF-8?q?=E6=9D=82=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0216.组合总和III.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md index f631c3cd..f08d77ea 100644 --- a/problems/0216.组合总和III.md +++ b/problems/0216.组合总和III.md @@ -235,6 +235,8 @@ public: } }; ``` +* 时间复杂度: O(n * 2^n) +* 空间复杂度: O(n) # 总结 From 025854e1678ba1e77a271b35a60159c1ad92ffbb Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 16:02:29 +0800 Subject: [PATCH 033/154] =?UTF-8?q?update=20=20=200017.=E7=94=B5=E8=AF=9D?= =?UTF-8?q?=E5=8F=B7=E7=A0=81=E7=9A=84=E5=AD=97=E6=AF=8D=E7=BB=84=E5=90=88?= =?UTF-8?q?=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82=E5=BA=A6=E5=88=86?= =?UTF-8?q?=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0017.电话号码的字母组合.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0017.电话号码的字母组合.md b/problems/0017.电话号码的字母组合.md index d1135497..d506bb88 100644 --- a/problems/0017.电话号码的字母组合.md +++ b/problems/0017.电话号码的字母组合.md @@ -183,6 +183,8 @@ public: } }; ``` +* 时间复杂度: O(3^m * 4^n)，其中 m 是对应四个字母的数字个数，n 是对应三个字母的数字个数 +* 空间复杂度: O(3^m * 4^n) 一些写法，是把回溯的过程放在递归函数里了，例如如下代码，我可以写成这样：（注意注释中不一样的地方） From d3290c7e773f66514625e8d678ff7695ae3f0f99 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 16:07:21 +0800 Subject: [PATCH 034/154] =?UTF-8?q?update=20=20=200039.=E7=BB=84=E5=90=88?= =?UTF-8?q?=E6=80=BB=E5=92=8C=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82?= =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0039.组合总和.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0039.组合总和.md b/problems/0039.组合总和.md index c4ee5ca6..e1e51923 100644 --- a/problems/0039.组合总和.md +++ b/problems/0039.组合总和.md @@ -214,6 +214,8 @@ public: } }; ``` +* 时间复杂度: O(n * 2^n)，注意这只是复杂度的上界，因为剪枝的存在，真实的时间复杂度远小于此 +* 空间复杂度: O(target) # 总结 From d68a1f633e532621ebd87578ae7c8daae4e5c36c Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 16:08:26 +0800 Subject: [PATCH 035/154] =?UTF-8?q?update=20=20=200040.=E7=BB=84=E5=90=88?= =?UTF-8?q?=E6=80=BB=E5=92=8CII=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D?= =?UTF-8?q?=E6=9D=82=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0040.组合总和II.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0040.组合总和II.md b/problems/0040.组合总和II.md index 83708df7..b708650a 100644 --- a/problems/0040.组合总和II.md +++ b/problems/0040.组合总和II.md @@ -214,6 +214,8 @@ public: }; ``` +* 时间复杂度: O(n * 2^n) +* 空间复杂度: O(n) ## 补充 From 65f059ac7a7168a8848bb7106f1050ce28afa862 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 16:27:02 +0800 Subject: [PATCH 036/154] =?UTF-8?q?update=20=200131.=E5=88=86=E5=89=B2?= =?UTF-8?q?=E5=9B=9E=E6=96=87=E4=B8=B2=20=EF=BC=9A=E6=B7=BB=E5=8A=A0?= =?UTF-8?q?=E5=A4=8D=E6=9D=82=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0131.分割回文串.md | 3 +++ 1 file changed, 3 insertions(+) diff --git a/problems/0131.分割回文串.md b/problems/0131.分割回文串.md index 30bba455..dfec7853 100644 --- a/problems/0131.分割回文串.md +++ b/problems/0131.分割回文串.md @@ -209,6 +209,9 @@ public: } }; ``` +* 时间复杂度: O(n * 2^n) +* 空间复杂度: O(n^2) + # 优化 上面的代码还存在一定的优化空间, 在于如何更高效的计算一个子字符串是否是回文字串。上述代码```isPalindrome```函数运用双指针的方法来判定对于一个字符串```s```, 给定起始下标和终止下标, 截取出的子字符串是否是回文字串。但是其中有一定的重复计算存在: From 33d5a6878e73800cae191d98276729c1ebb77860 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 16:33:52 +0800 Subject: [PATCH 037/154] =?UTF-8?q?=20update=20=20=200078.=E5=AD=90?= =?UTF-8?q?=E9=9B=86=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82=E5=BA=A6?= =?UTF-8?q?=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0078.子集.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0078.子集.md b/problems/0078.子集.md index f26d0821..07418fbc 100644 --- a/problems/0078.子集.md +++ b/problems/0078.子集.md @@ -149,6 +149,8 @@ public: }; ``` +* 时间复杂度: O(n * 2^n) +* 空间复杂度: O(n) 在注释中，可以发现可以不写终止条件，因为本来我们就要遍历整棵树。 From 3accc7602597a2d6c7af017326d8caf9cb30be16 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 16:34:55 +0800 Subject: [PATCH 038/154] =?UTF-8?q?update=20=20=200090.=E5=AD=90=E9=9B=86I?= =?UTF-8?q?I=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82=E5=BA=A6?= =?UTF-8?q?=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0090.子集II.md | 3 +++ 1 file changed, 3 insertions(+) diff --git a/problems/0090.子集II.md b/problems/0090.子集II.md index 1a9f8fda..63f75d29 100644 --- a/problems/0090.子集II.md +++ b/problems/0090.子集II.md @@ -83,6 +83,9 @@ public: } }; ``` +* 时间复杂度: O(n * 2^n) +* 空间复杂度: O(n) + 使用set去重的版本。 ```CPP From 09975d3d380c3151538270fe8cac25d39a476281 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 16:37:53 +0800 Subject: [PATCH 039/154] =?UTF-8?q?update=20=20=200491.=E9=80=92=E5=A2=9E?= =?UTF-8?q?=E5=AD=90=E5=BA=8F=E5=88=97=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D?= =?UTF-8?q?=E6=9D=82=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0491.递增子序列.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0491.递增子序列.md b/problems/0491.递增子序列.md index 436dbf01..d6c6b9c9 100644 --- a/problems/0491.递增子序列.md +++ b/problems/0491.递增子序列.md @@ -139,6 +139,8 @@ public: } }; ``` +* 时间复杂度: O(n * 2^n) +* 空间复杂度: O(n) ## 优化 From 9d581e86fe858577263597ec5b0a1aa5dda24c2e Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 16:38:59 +0800 Subject: [PATCH 040/154] =?UTF-8?q?update=20=200046.=E5=85=A8=E6=8E=92?= =?UTF-8?q?=E5=88=97=20=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82?= =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0046.全排列.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0046.全排列.md b/problems/0046.全排列.md index e08aec94..fb70be41 100644 --- a/problems/0046.全排列.md +++ b/problems/0046.全排列.md @@ -136,6 +136,8 @@ public: } }; ``` +* 时间复杂度: O(n!) +* 空间复杂度: O(n) ## 总结 From 1a9cb629ae87aa43a7eac4dea0a6732492a8a242 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 16:39:37 +0800 Subject: [PATCH 041/154] =?UTF-8?q?update=20=200047.=E5=85=A8=E6=8E=92?= =?UTF-8?q?=E5=88=97II=20=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82?= =?UTF-8?q?=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0047.全排列II.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0047.全排列II.md b/problems/0047.全排列II.md index b1908fb4..3ff3fb8f 100644 --- a/problems/0047.全排列II.md +++ b/problems/0047.全排列II.md @@ -99,6 +99,8 @@ public: }; ``` +* 时间复杂度: O(n) +* 空间复杂度: O(n) ## 拓展 From 2a7f4d5d4b4107c0aa85960e0e46adb56d665def Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 16:44:28 +0800 Subject: [PATCH 042/154] =?UTF-8?q?update=20=20=200051.N=E7=9A=87=E5=90=8E?= =?UTF-8?q?=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D=E6=9D=82=E5=BA=A6=E5=88=86?= =?UTF-8?q?=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0051.N皇后.md | 3 +++ 1 file changed, 3 insertions(+) diff --git a/problems/0051.N皇后.md b/problems/0051.N皇后.md index bd1d1c9b..54580cf7 100644 --- a/problems/0051.N皇后.md +++ b/problems/0051.N皇后.md @@ -208,6 +208,9 @@ public: } }; ``` +* 时间复杂度: O(n!) +* 空间复杂度: O(n) + 可以看出，除了验证棋盘合法性的代码，省下来部分就是按照回溯法模板来的。 From 7f699b59167263645112138b8ff4a99e8c606846 Mon Sep 17 00:00:00 2001 From: Yuhao Ju Date: Thu, 27 Apr 2023 17:15:12 +0800 Subject: [PATCH 043/154] =?UTF-8?q?update=20=20=200093.=E5=A4=8D=E5=8E=9FI?= =?UTF-8?q?P=E5=9C=B0=E5=9D=80=EF=BC=9A=E6=B7=BB=E5=8A=A0=E5=A4=8D?= =?UTF-8?q?=E6=9D=82=E5=BA=A6=E5=88=86=E6=9E=90?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0093.复原IP地址.md | 2 ++ 1 file changed, 2 insertions(+) diff --git a/problems/0093.复原IP地址.md b/problems/0093.复原IP地址.md index 9d4d5918..161fb96e 100644 --- a/problems/0093.复原IP地址.md +++ b/problems/0093.复原IP地址.md @@ -244,6 +244,8 @@ public: }; ``` +* 时间复杂度: O(3^4)，IP地址最多包含4个数字，每个数字最多有3种可能的分割方式，则搜索树的最大深度为4，每个节点最多有3个子节点。 +* 空间复杂度: O(n) # 总结 From 42cdaa2e9a2639dab6931a5733aabf6ed92df62a Mon Sep 17 00:00:00 2001 From: blockChain-Fans <33158355+1055373165@users.noreply.github.com> Date: Thu, 27 Apr 2023 23:34:50 +0800 Subject: [PATCH 044/154] =?UTF-8?q?Update=200035.=E6=90=9C=E7=B4=A2?= =?UTF-8?q?=E6=8F=92=E5=85=A5=E4=BD=8D=E7=BD=AE.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 返回值应该是 right+1 把 --- problems/0035.搜索插入位置.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0035.搜索插入位置.md b/problems/0035.搜索插入位置.md index 58340c21..efc87577 100644 --- a/problems/0035.搜索插入位置.md +++ b/problems/0035.搜索插入位置.md @@ -274,7 +274,7 @@ func searchInsert(nums []int, target int) int { left = mid + 1 } } - return len(nums) + return right+1 } ``` From 31d2755d18a216384495f25654cd4510c360dadf Mon Sep 17 00:00:00 2001 From: xiaodi007 <334830452@qq.com> Date: Fri, 28 Apr 2023 11:17:08 +0800 Subject: [PATCH 045/154] fix 0343 py code --- problems/0343.整数拆分.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0343.整数拆分.md b/problems/0343.整数拆分.md index c2fbbdde..812bf67b 100644 --- a/problems/0343.整数拆分.md +++ b/problems/0343.整数拆分.md @@ -254,7 +254,7 @@ class Solution: # 假设对正整数 i 拆分出的第一个正整数是 j（1 <= j < i），则有以下两种方案： # 1) 将 i 拆分成 j 和 i−j 的和，且 i−j 不再拆分成多个正整数，此时的乘积是 j * (i-j) # 2) 将 i 拆分成 j 和 i−j 的和，且 i−j 继续拆分成多个正整数，此时的乘积是 j * dp[i-j] - for j in range(1, i / 2 + 1): + for j in range(1, i // 2 + 1): dp[i] = max(dp[i], max(j * (i - j), j * dp[i - j])) return dp[n] ``` From 14c25f2eeff8e024f822485c4fbdfddcb0c3710e Mon Sep 17 00:00:00 2001 From: fwqaaq Date: Fri, 28 Apr 2023 18:30:16 +0800 Subject: [PATCH 046/154] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E7=90=86?= =?UTF-8?q?=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85-1.md=20about=20?= =?UTF-8?q?rust?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-1.md | 33 ++++++++++++++++++++++++ 1 file changed, 33 insertions(+) diff --git a/problems/背包理论基础01背包-1.md b/problems/背包理论基础01背包-1.md index c45fc3d3..c2525248 100644 --- a/problems/背包理论基础01背包-1.md +++ b/problems/背包理论基础01背包-1.md @@ -573,6 +573,39 @@ object Solution { } ``` +### Rust + +```rust +pub struct Solution; + +impl Solution { + pub fn wei_bag_problem1(weight: Vec, value: Vec, bag_size: usize) -> usize { + let mut dp = vec![vec![0; bag_size + 1]; weight.len()]; + for j in weight[0]..=weight.len() { + dp[0][j] = value[0]; + } + + for i in 1..weight.len() { + for j in 0..=bag_size { + match j < weight[i] { + true => dp[i][j] = dp[i - 1][j], + false => dp[i][j] = dp[i - 1][j].max(dp[i - 1][j - weight[i]] + value[i]), + } + } + } + dp[weight.len() - 1][bag_size] + } +} + +#[test] +fn test_wei_bag_problem1() { + println!( + "{}", + Solution::wei_bag_problem1(vec![1, 3, 4], vec![15, 20, 30], 4) + ); +} +``` +

From e6ad637e6475c73766f401b052a1c523d171d059 Mon Sep 17 00:00:00 2001 From: fwqaaq Date: Fri, 28 Apr 2023 18:50:56 +0800 Subject: [PATCH 047/154] =?UTF-8?q?Update=20=E8=83=8C=E5=8C=85=E7=90=86?= =?UTF-8?q?=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85-2.md=20about=20?= =?UTF-8?q?rust?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-2.md | 28 ++++++++++++++++++++++++ 1 file changed, 28 insertions(+) diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md index baa4107f..1ce90440 100644 --- a/problems/背包理论基础01背包-2.md +++ b/problems/背包理论基础01背包-2.md @@ -406,6 +406,34 @@ object Solution { } ``` +### Rust + +```rust +pub struct Solution; + +impl Solution { + pub fn wei_bag_problem2(weight: Vec, value: Vec, bag_size: usize) -> usize { + let mut dp = vec![0; bag_size + 1]; + for i in 0..weight.len() { + for j in (weight[i]..=bag_size).rev() { + if j >= weight[i] { + dp[j] = dp[j].max(dp[j - weight[i]] + value[i]); + } + } + } + dp[dp.len() - 1] + } +} + +#[test] +fn test_wei_bag_problem2() { + println!( + "{}", + Solution::wei_bag_problem1(vec![1, 3, 4], vec![15, 20, 30], 4) + ); +} +``` +

From d2650cc436980c2553003e3c7bdba2e8579aa53b Mon Sep 17 00:00:00 2001 From: fwqaaq Date: Fri, 28 Apr 2023 18:51:38 +0800 Subject: [PATCH 048/154] =?UTF-8?q?Update=20problems/=E8=83=8C=E5=8C=85?= =?UTF-8?q?=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=8001=E8=83=8C=E5=8C=85-2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/背包理论基础01背包-2.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/背包理论基础01背包-2.md b/problems/背包理论基础01背包-2.md index 1ce90440..3b798334 100644 --- a/problems/背包理论基础01背包-2.md +++ b/problems/背包理论基础01背包-2.md @@ -429,7 +429,7 @@ impl Solution { fn test_wei_bag_problem2() { println!( "{}", - Solution::wei_bag_problem1(vec![1, 3, 4], vec![15, 20, 30], 4) + Solution::wei_bag_problem2(vec![1, 3, 4], vec![15, 20, 30], 4) ); } ``` From 3ff604055147700bd66ad8c83b43ee6f073e4265 Mon Sep 17 00:00:00 2001 From: lzxzz <1042183935@qq.com> Date: Sat, 29 Apr 2023 09:49:25 +0800 Subject: [PATCH 049/154] optimize --- ....删除字符串中的所有相邻重复项.md | 15 ++++++++------- 1 file changed, 8 insertions(+), 7 deletions(-) diff --git a/problems/1047.删除字符串中的所有相邻重复项.md b/problems/1047.删除字符串中的所有相邻重复项.md index 694f1a92..486d198b 100644 --- a/problems/1047.删除字符串中的所有相邻重复项.md +++ b/problems/1047.删除字符串中的所有相邻重复项.md @@ -264,14 +264,15 @@ javaScript: ```js var removeDuplicates = function(s) { - const stack = []; - for(const x of s) { - let c = null; - if(stack.length && x === (c = stack.pop())) continue; - c && stack.push(c); - stack.push(x); + const result = [] + for(const i of s){ + if(i === result[result.length-1]){ + result.pop() + }else{ + result.push(i) + } } - return stack.join(""); + return result.join('') }; ``` From 76d347076b1ff05769c6bbb8844027363608e017 Mon Sep 17 00:00:00 2001 From: fwqaaq Date: Sat, 29 Apr 2023 10:27:48 +0800 Subject: [PATCH 050/154] =?UTF-8?q?Update=200416.=E5=88=86=E5=89=B2?= =?UTF-8?q?=E7=AD=89=E5=92=8C=E5=AD=90=E9=9B=86.md=20about=20rust?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 去除不必要的方法 --- problems/0416.分割等和子集.md | 23 ++++++++++------------- 1 file changed, 10 insertions(+), 13 deletions(-) diff --git a/problems/0416.分割等和子集.md b/problems/0416.分割等和子集.md index 8726bf95..8115e18e 100644 --- a/problems/0416.分割等和子集.md +++ b/problems/0416.分割等和子集.md @@ -406,24 +406,21 @@ var canPartition = function(nums) { ```Rust impl Solution { - fn max(a: usize, b: usize) -> usize { - if a > b { a } else { b } - } pub fn can_partition(nums: Vec) -> bool { - let nums = nums.iter().map(|x| *x as usize).collect::>(); - let mut sum = 0; - let mut dp: Vec = vec![0; 10001]; - for i in 0..nums.len() { - sum += nums[i]; + let sum = nums.iter().sum::() as usize; + if sum % 2 == 1 { + return false; } - if sum % 2 == 1 { return false; } let target = sum / 2; - for i in 0..nums.len() { - for j in (nums[i]..=target).rev() { - dp[j] = Self::max(dp[j], dp[j - nums[i]] + nums[i]); + let mut dp = vec![0; target + 1]; + for n in nums { + for j in (n as usize..=target).rev() { + dp[j] = dp[j].max(dp[j - n as usize] + n) } } - if dp[target] == target { return true; } + if dp[target] == target as i32 { + return true; + } false } } From da4116bcb60b3bf6efa7f44e8e20877e987fadfe Mon Sep 17 00:00:00 2001 From: Lozakaka <102352821+Lozakaka@users.noreply.github.com> Date: Sat, 29 Apr 2023 05:45:01 -0400 Subject: [PATCH 051/154] =?UTF-8?q?=E6=96=B0=E5=A2=9E=20Java=20=E7=B5=B1?= =?UTF-8?q?=E4=B8=80=E8=BF=AD=E4=BB=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 新增 Java 統一迭代法 --- problems/0501.二叉搜索树中的众数.md | 53 ++++++++++++++++++++ 1 file changed, 53 insertions(+) diff --git a/problems/0501.二叉搜索树中的众数.md b/problems/0501.二叉搜索树中的众数.md index b7ef606f..e467bf8a 100644 --- a/problems/0501.二叉搜索树中的众数.md +++ b/problems/0501.二叉搜索树中的众数.md @@ -472,6 +472,59 @@ class Solution { } } ``` +統一迭代法 +```Java +class Solution { + public int[] findMode(TreeNode root) { + int count = 0; + int maxCount = 0; + TreeNode pre = null; + LinkedList res = new LinkedList<>(); + Stack stack = new Stack<>(); + + if(root != null) + stack.add(root); + + while(!stack.isEmpty()){ + TreeNode curr = stack.peek(); + if(curr != null){ + stack.pop(); + if(curr.right != null) + stack.add(curr.right); + stack.add(curr); + stack.add(null); + if(curr.left != null) + stack.add(curr.left); + }else{ + stack.pop(); + TreeNode temp = stack.pop(); + if(pre == null) + count = 1; + else if(pre != null && pre.val == temp.val) + count++; + else + count = 1; + pre = temp; + if(count == maxCount) + res.add(temp.val); + if(count > maxCount){ + maxCount = count; + res.clear(); + res.add(temp.val); + } + } + } + int[] result = new int[res.size()]; + int i = 0; + for (int x : res){ + result[i] = x; + i++; + } + return result; + } +} +``` + ## Python From 513707ca713b2b199f8202e953b1a8dd2c40d447 Mon Sep 17 00:00:00 2001 From: skyclouds2001 <95597335+skyclouds2001@users.noreply.github.com> Date: Sun, 30 Apr 2023 14:24:26 +0800 Subject: [PATCH 052/154] =?UTF-8?q?=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4?= =?UTF-8?q?=E6=96=87=E6=A1=A3=E7=BB=93=E6=9E=84=E6=A0=BC=E5=BC=8F=E5=8C=96?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/面试题02.07.链表相交.md | 11 ++++++----- 1 file changed, 6 insertions(+), 5 deletions(-) diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index 30f5c467..9e0c29f5 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -101,8 +101,8 @@ public: ## 其他语言版本 +Java： -### Java ```Java public class Solution { public ListNode getIntersectionNode(ListNode headA, ListNode headB) { @@ -150,9 +150,9 @@ public class Solution { } ``` -### Python -```python +Python： +```python class Solution: def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: lenA, lenB = 0, 0 @@ -179,7 +179,7 @@ class Solution: return None ``` -### Go +Go： ```go func getIntersectionNode(headA, headB *ListNode) *ListNode { @@ -240,7 +240,7 @@ func getIntersectionNode(headA, headB *ListNode) *ListNode { } ``` -### javaScript +JavaScript： ```js var getListLen = function(head) { @@ -352,6 +352,7 @@ ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) { ``` Scala: + ```scala object Solution { def getIntersectionNode(headA: ListNode, headB: ListNode): ListNode = { From 113d1192a92b50ad8154b4b7717cdb5d9186d7a0 Mon Sep 17 00:00:00 2001 From: Mrzhugq <84071063+Mrzhugq@users.noreply.github.com> Date: Sun, 30 Apr 2023 16:45:31 +0800 Subject: [PATCH 053/154] =?UTF-8?q?Update=200035.=E6=90=9C=E7=B4=A2?= =?UTF-8?q?=E6=8F=92=E5=85=A5=E4=BD=8D=E7=BD=AE.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 195行把空间复杂度写成时间复杂度了 --- problems/0035.搜索插入位置.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/0035.搜索插入位置.md b/problems/0035.搜索插入位置.md index 58340c21..a37d67f5 100644 --- a/problems/0035.搜索插入位置.md +++ b/problems/0035.搜索插入位置.md @@ -191,8 +191,8 @@ public: }; ``` -* 时间复杂度：$O(\log n)$ -* 时间复杂度：$O(1)$ +* 时间复杂度：O(log n) +* 空间复杂度：O(1) ## 总结 From 18a16f9faffb7b796d174fb150e3dd601f6f82f4 Mon Sep 17 00:00:00 2001 From: Jia Tan Date: Sun, 30 Apr 2023 20:24:32 +0000 Subject: [PATCH 054/154] Ignore .DS_Store files. --- .DS_Store | Bin 6148 -> 0 bytes .gitignore | 1 + 2 files changed, 1 insertion(+) delete mode 100644 .DS_Store create mode 100644 .gitignore diff --git a/.DS_Store b/.DS_Store deleted file mode 100644 index da03c1c1b207c3ee04079084ecb15dcfff97ef95..0000000000000000000000000000000000000000 GIT binary patch literal 0 HcmV?d00001 literal 6148 zcmeHK%}T>S5T317Q;OJwLXQhx3-<3)^bl)(0V4`psf{T$m}X1UTBHSVD*@yHz(SRn%3w1`N_?k*PTX z=s*u5cO3N-(d77z49It9K_3Pnz=e13&kcl&V-IdW@S|a|_{35R>BXgGloVwt;#QN%@Vis8VsHj`Xe9O(bW6^; z=;?2>^71?{rY9N=vv=!y_4uZF|F|A+0!!OY6cR47cXj`MRB4$Rg_Eiph0d}LrjcXRUo z-}}7&|CmHQVt^R Date: Sun, 30 Apr 2023 13:40:16 -0700 Subject: [PATCH 055/154] =?UTF-8?q?Update=20=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=80.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树理论基础.md | 27 ++++++++++++++------------- 1 file changed, 14 insertions(+), 13 deletions(-) diff --git a/problems/二叉树理论基础.md b/problems/二叉树理论基础.md index 3afedc82..dbf42ca4 100644 --- a/problems/二叉树理论基础.md +++ b/problems/二叉树理论基础.md @@ -195,15 +195,16 @@ Java： ```java public class TreeNode { int val; - TreeNode left; - TreeNode right; - TreeNode() {} - TreeNode(int val) { this.val = val; } - TreeNode(int val, TreeNode left, TreeNode right) { - this.val = val; - this.left = left; - this.right = right; - } + TreeNode left; + TreeNode right; + + TreeNode() {} + TreeNode(int val) { this.val = val; } + TreeNode(int val, TreeNode left, TreeNode right) { + this.val = val; + this.left = left; + this.right = right; + } } ``` @@ -212,10 +213,10 @@ Python： ```python class TreeNode: - def __init__(self, value): - self.value = value - self.left = None - self.right = None + def __init__(self, val, left = None, right = None): + self.val = val + self.left = left + self.right = right ``` Go： From 9617a8b3fedc3996645977df31104159864b3462 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 1 May 2023 15:46:15 -0500 Subject: [PATCH 056/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 75 +++++++++++++---------- 1 file changed, 44 insertions(+), 31 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 26f0ae2d..e777c2a4 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -171,47 +171,60 @@ python3代码： ```python +# 利用长度法 +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +from collections import deque class Solution: - """二叉树层序遍历迭代解法""" - - def levelOrder(self, root: TreeNode) -> List[List[int]]: - results = [] + def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: if not root: - return results - - from collections import deque - que = deque([root]) - - while que: - size = len(que) - result = [] - for _ in range(size): - cur = que.popleft() - result.append(cur.val) + return [] + queue = deque([root]) + result = [] + while queue: + level = [] + for _ in range(len(queue)): + cur = queue.popleft() + level.append(cur.val) if cur.left: - que.append(cur.left) + queue.append(cur.left) if cur.right: - que.append(cur.right) - results.append(result) - - return results + queue.append(cur.right) + result.append(level) + return result ``` - ```python # 递归法 +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: - def levelOrder(self, root: TreeNode) -> List[List[int]]: - res = [] - def helper(root, depth): - if not root: return [] - if len(res) == depth: res.append([]) # start the current depth - res[depth].append(root.val) # fulfil the current depth - if root.left: helper(root.left, depth + 1) # process child nodes for the next depth - if root.right: helper(root.right, depth + 1) - helper(root, 0) - return res + def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: + levels = [] + self.helper(root, 0, levels) + return levels + + def helper(self, node, level, levels): + if not node: + return + if len(levels) == level: + levels.append([]) + levels[level].append(node.val) + self.helper(node.left, level + 1, levels) + self.helper(node.right, level + 1, levels) + + ``` + + go: ```go From f80761a43833de34dc355d48f729c64ecef63afb Mon Sep 17 00:00:00 2001 From: JaneyLin <105125897+janeyziqinglin@users.noreply.github.com> Date: Tue, 2 May 2023 23:34:57 -0400 Subject: [PATCH 057/154] =?UTF-8?q?Update=200300.=E6=9C=80=E9=95=BF?= =?UTF-8?q?=E4=B8=8A=E5=8D=87=E5=AD=90=E5=BA=8F=E5=88=97.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit result should be initialized to 1. --- problems/0300.最长上升子序列.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0300.最长上升子序列.md b/problems/0300.最长上升子序列.md index e8cb0b5f..01d34949 100644 --- a/problems/0300.最长上升子序列.md +++ b/problems/0300.最长上升子序列.md @@ -149,7 +149,7 @@ class Solution: if len(nums) <= 1: return len(nums) dp = [1] * len(nums) - result = 0 + result = 1 for i in range(1, len(nums)): for j in range(0, i): if nums[i] > nums[j]: From c76ed37fc487bc7da34403bbb484e9e42b582efb Mon Sep 17 00:00:00 2001 From: Lozakaka <102352821+Lozakaka@users.noreply.github.com> Date: Tue, 2 May 2023 23:55:58 -0400 Subject: [PATCH 058/154] adding java iteration method MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 新增java 迭代方法 --- problems/0669.修剪二叉搜索树.md | 42 ++++++++++++++++++++++++++ 1 file changed, 42 insertions(+) diff --git a/problems/0669.修剪二叉搜索树.md b/problems/0669.修剪二叉搜索树.md index 18d8a0cc..5739f762 100644 --- a/problems/0669.修剪二叉搜索树.md +++ b/problems/0669.修剪二叉搜索树.md @@ -248,6 +248,8 @@ public: ## Java +**递归** + ```Java class Solution { public TreeNode trimBST(TreeNode root, int low, int high) { @@ -269,6 +271,46 @@ class Solution { ``` +**迭代** + +```Java +class Solution { + //iteration + public TreeNode trimBST(TreeNode root, int low, int high) { + if(root == null) + return null; + while(root != null && (root.val < low || root.val > high)){ + if(root.val < low) + root = root.right; + else + root = root.left; + } + + TreeNode curr = root; + + //deal with root's left sub-tree, and deal with the value smaller than low. + while(curr != null){ + while(curr.left != null && curr.left.val < low){ + curr.left = curr.left.right; + } + curr = curr.left; + } + //go back to root; + curr = root; + + //deal with root's righg sub-tree, and deal with the value bigger than high. + while(curr != null){ + while(curr.right != null && curr.right.val > high){ + curr.right = curr.right.left; + } + curr = curr.right; + } + return root; + } +} + +```` + ## Python **递归** From 424d5c224f229aaee353b3bbae8996b8e5dcbef2 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 15:06:11 -0500 Subject: [PATCH 059/154] =?UTF-8?q?Update=200027.=E7=A7=BB=E9=99=A4?= =?UTF-8?q?=E5=85=83=E7=B4=A0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0027.移除元素.md | 17 ++++++++++++++++- 1 file changed, 16 insertions(+), 1 deletion(-) diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 91150c74..6dc6404e 100644 --- a/problems/0027.移除元素.md +++ b/problems/0027.移除元素.md @@ -198,6 +198,7 @@ Python： ``` python 3 +（版本一）快慢指针法 class Solution: def removeElement(self, nums: List[int], val: int) -> int: # 快慢指针 @@ -213,7 +214,21 @@ class Solution: return slow ``` - +``` python 3 +（版本二）暴力递归法 +class Solution: + def removeElement(self, nums: List[int], val: int) -> int: + i, l = 0, len(nums) + while i < l: + if nums[i] == val: # 找到等于目标值的节点 + for j in range(i+1, l): # 移除该元素，并将后面元素向前平移 + nums[j - 1] = nums[j] + l -= 1 + i -= 1 + i += 1 + return l + +``` Go： From 53fbfc8339a6b61cb708ec87356e764db0f2b1ba Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 15:08:17 -0500 Subject: [PATCH 060/154] =?UTF-8?q?Update=200027.=E7=A7=BB=E9=99=A4?= =?UTF-8?q?=E5=85=83=E7=B4=A0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0027.移除元素.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0027.移除元素.md b/problems/0027.移除元素.md index 6dc6404e..90801153 100644 --- a/problems/0027.移除元素.md +++ b/problems/0027.移除元素.md @@ -215,7 +215,7 @@ class Solution: ``` ``` python 3 -（版本二）暴力递归法 +（版本二）暴力法 class Solution: def removeElement(self, nums: List[int], val: int) -> int: i, l = 0, len(nums) From 934dd4e3131485802edd5d7c45fd60dd4f8f2827 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 15:19:23 -0500 Subject: [PATCH 061/154] =?UTF-8?q?Update=200977.=E6=9C=89=E5=BA=8F?= =?UTF-8?q?=E6=95=B0=E7=BB=84=E7=9A=84=E5=B9=B3=E6=96=B9.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0977.有序数组的平方.md | 41 ++++++++++++++++++-------- 1 file changed, 28 insertions(+), 13 deletions(-) diff --git a/problems/0977.有序数组的平方.md b/problems/0977.有序数组的平方.md index 57f8de02..4fbdd1cd 100644 --- a/problems/0977.有序数组的平方.md +++ b/problems/0977.有序数组的平方.md @@ -140,22 +140,37 @@ class Solution { Python： ```Python +（版本一）双指针法 class Solution: def sortedSquares(self, nums: List[int]) -> List[int]: - n = len(nums) - i,j,k = 0,n - 1,n - 1 - ans = [-1] * n - while i <= j: - lm = nums[i] ** 2 - rm = nums[j] ** 2 - if lm > rm: - ans[k] = lm - i += 1 + l, r, i = 0, len(nums)-1, len(nums)-1 + res = [float('inf')] * len(nums) # 需要提前定义列表，存放结果 + while l <= r: + if nums[l] ** 2 < nums[r] ** 2: # 左右边界进行对比，找出最大值 + res[i] = nums[r] ** 2 + r -= 1 # 右指针往左移动 else: - ans[k] = rm - j -= 1 - k -= 1 - return ans + res[i] = nums[l] ** 2 + l += 1 # 左指针往右移动 + i -= 1 # 存放结果的指针需要往前平移一位 + return res +``` + +```Python +（版本二）暴力排序法 +class Solution: + def sortedSquares(self, nums: List[int]) -> List[int]: + for i in range(len(nums)): + nums[i] *= nums[i] + nums.sort() + return nums +``` + +```Python +（版本三）暴力排序法+列表推导法 +class Solution: + def sortedSquares(self, nums: List[int]) -> List[int]: + return sorted(x*x for x in nums) ``` Go： From 5f65e3ba24617dc81b199bcbb356f47517f7fbee Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 15:32:36 -0500 Subject: [PATCH 062/154] =?UTF-8?q?Update=200209.=E9=95=BF=E5=BA=A6?= =?UTF-8?q?=E6=9C=80=E5=B0=8F=E7=9A=84=E5=AD=90=E6=95=B0=E7=BB=84.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0209.长度最小的子数组.md | 46 ++++++++++++++++++----- 1 file changed, 36 insertions(+), 10 deletions(-) diff --git a/problems/0209.长度最小的子数组.md b/problems/0209.长度最小的子数组.md index 5a8f91af..d7ae4780 100644 --- a/problems/0209.长度最小的子数组.md +++ b/problems/0209.长度最小的子数组.md @@ -173,18 +173,44 @@ class Solution { Python： ```python +（版本一）滑动窗口法 class Solution: def minSubArrayLen(self, s: int, nums: List[int]) -> int: - res = float("inf") # 定义一个无限大的数 - Sum = 0 # 滑动窗口数值之和 - i = 0 # 滑动窗口起始位置 - for j in range(len(nums)): - Sum += nums[j] - while Sum >= s: - res = min(res, j-i+1) - Sum -= nums[i] - i += 1 - return 0 if res == float("inf") else res + l = len(nums) + left = 0 + right = 0 + min_len = float('inf') + cur_sum = 0 #当前的累加值 + + while right < l: + cur_sum += nums[right] + + while cur_sum >= s: # 当前累加值大于目标值 + min_len = min(min_len, right - left + 1) + cur_sum -= nums[left] + left += 1 + + right += 1 + + return min_len if min_len != float('inf') else 0 +``` + +```python +（版本二）暴力法 +class Solution: + def minSubArrayLen(self, s: int, nums: List[int]) -> int: + l = len(nums) + min_len = float('inf') + + for i in range(l): + cur_sum = 0 + for j in range(i, l): + cur_sum += nums[j] + if cur_sum >= s: + min_len = min(min_len, j - i + 1) + break + + return min_len if min_len != float('inf') else 0 ``` Go： From f45b1f1d28a668e2c90ad66ac0d2141d59fc73ae Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 16:19:13 -0500 Subject: [PATCH 063/154] =?UTF-8?q?Update=200203.=E7=A7=BB=E9=99=A4?= =?UTF-8?q?=E9=93=BE=E8=A1=A8=E5=85=83=E7=B4=A0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0203.移除链表元素.md | 20 +++++++++++++------- 1 file changed, 13 insertions(+), 7 deletions(-) diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index b52f16ea..6a0de282 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -307,21 +307,27 @@ public ListNode removeElements(ListNode head, int val) { Python： ```python +（版本一）虚拟头节点法 # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): # self.val = val # self.next = next class Solution: - def removeElements(self, head: ListNode, val: int) -> ListNode: - dummy_head = ListNode(next=head) #添加一个虚拟节点 - cur = dummy_head - while cur.next: - if cur.next.val == val: - cur.next = cur.next.next #删除cur.next节点 + def removeElements(self, head: Optional[ListNode], val: int) -> Optional[ListNode]: + # 创建虚拟头部节点以简化删除过程 + dummy_head = ListNode(next = head) + + # 遍历列表并删除值为val的节点 + current = dummy_head + while current.next: + if current.next.val == val: + current.next = current.next.next else: - cur = cur.next + current = current.next + return dummy_head.next + ``` Go： From 70d8379ff7fe99edbbd2d006b9b3d9a5068c66fc Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 16:26:16 -0500 Subject: [PATCH 064/154] =?UTF-8?q?Update=200707.=E8=AE=BE=E8=AE=A1?= =?UTF-8?q?=E9=93=BE=E8=A1=A8.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0707.设计链表.md | 294 +++++++++++++++++----------------- 1 file changed, 144 insertions(+), 150 deletions(-) diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index c72b7327..aa04d0e1 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -485,172 +485,166 @@ class MyLinkedList { Python： ```python -# 单链表 -class MyLinkedList1: - - def __init__(self): - self.dummy_head = Node()# 添加虚拟头指针，便于操作 - self.size = 0 # 设置一个链表长度的属性，便于后续操作，注意每次增和删的时候都要更新 - - def get(self, index): - """ - :type index: int - :rtype: int - """ - if index < 0 or index >= self.size: - return -1 - cur = self.dummy_head.next - while(index): - cur = cur.next - index -= 1 - return cur.val - - def addAtHead(self, val): - """ - :type val: int - :rtype: None - """ - new_node = Node(val) - new_node.next = self.dummy_head.next - self.dummy_head.next = new_node - self.size += 1 - - def addAtTail(self, val): - """ - :type val: int - :rtype: None - """ - new_node = Node(val) - cur = self.dummy_head - while(cur.next): - cur = cur.next - cur.next = new_node - self.size += 1 - - def addAtIndex(self, index, val): - """ - :type index: int - :type val: int - :rtype: None - """ - if index < 0: - self.addAtHead(val) - return - elif index == self.size: - self.addAtTail(val) - return - elif index > self.size: - return - - node = Node(val) - cur = self.dummy_head - while(index): - cur = cur.next - index -= 1 - node.next = cur.next - cur.next = node - self.size += 1 - - def deleteAtIndex(self, index): - """ - :type index: int - :rtype: None - """ - if index < 0 or index >= self.size: - return - pre = self.dummy_head - while(index): - pre = pre.next - index -= 1 - pre.next = pre.next.next - self.size -= 1 - -# 双链表 -# 相对于单链表, Node新增了prev属性 -class Node: - - def __init__(self, val=0, next = None, prev = None): +（版本一）单链表法 +class ListNode: + def __init__(self, val=0, next=None): self.val = val self.next = next - self.prev = prev - + class MyLinkedList: - def __init__(self): - self._head, self._tail = Node(0), Node(0) # 虚拟节点 - self._head.next, self._tail.prev = self._tail, self._head - self._count = 0 # 添加的节点数 - - def _get_node(self, index: int) -> Node: - # 当index小于_count//2时, 使用_head查找更快, 反之_tail更快 - if index >= self._count // 2: - # 使用prev往前找 - node = self._tail - for _ in range(self._count - index): - node = node.prev - else: - # 使用next往后找 - node = self._head - for _ in range(index + 1): - node = node.next - return node - - - def _update(self, prev: Node, next: Node, val: int) -> None: - """ - 更新节点 - :param prev: 相对于更新的前一个节点 - :param next: 相对于更新的后一个节点 - :param val: 要添加的节点值 - """ - # 计数累加 - self._count += 1 - node = Node(val) - prev.next, next.prev = node, node - node.prev, node.next = prev, next + self.dummy_head = ListNode() + self.size = 0 def get(self, index: int) -> int: - """ - Get the value of the index-th node in the linked list. If the index is invalid, return -1. - """ - if 0 <= index < self._count: - node = self._get_node(index) - return node.val - else: + if index < 0 or index >= self.size: return -1 + + current = self.dummy_head.next + for i in range(index): + current = current.next + + return current.val def addAtHead(self, val: int) -> None: - """ - Add a node of value val before the first element of the linked list. After the insertion, the new node will be the first node of the linked list. - """ - self._update(self._head, self._head.next, val) + self.dummy_head.next = ListNode(val, self.dummy_head.next) + self.size += 1 def addAtTail(self, val: int) -> None: - """ - Append a node of value val to the last element of the linked list. - """ - self._update(self._tail.prev, self._tail, val) + current = self.dummy_head + while current.next: + current = current.next + current.next = ListNode(val) + self.size += 1 def addAtIndex(self, index: int, val: int) -> None: - """ - Add a node of value val before the index-th node in the linked list. If index equals to the length of linked list, the node will be appended to the end of linked list. If index is greater than the length, the node will not be inserted. - """ - if index < 0: - index = 0 - elif index > self._count: + if index < 0 or index > self.size: return - node = self._get_node(index) - self._update(node.prev, node, val) + + current = self.dummy_head + for i in range(index): + current = current.next + current.next = ListNode(val, current.next) + self.size += 1 def deleteAtIndex(self, index: int) -> None: - """ - Delete the index-th node in the linked list, if the index is valid. - """ - if 0 <= index < self._count: - node = self._get_node(index) - # 计数-1 - self._count -= 1 - node.prev.next, node.next.prev = node.next, node.prev + if index < 0 or index >= self.size: + return + + current = self.dummy_head + for i in range(index): + current = current.next + current.next = current.next.next + self.size -= 1 + + +# Your MyLinkedList object will be instantiated and called as such: +# obj = MyLinkedList() +# param_1 = obj.get(index) +# obj.addAtHead(val) +# obj.addAtTail(val) +# obj.addAtIndex(index,val) +# obj.deleteAtIndex(index) +``` + + +```python +（版本二）双链表法 +class ListNode: + def __init__(self, val=0, prev=None, next=None): + self.val = val + self.prev = prev + self.next = next + +class MyLinkedList: + def __init__(self): + self.head = None + self.tail = None + self.size = 0 + + def get(self, index: int) -> int: + if index < 0 or index >= self.size: + return -1 + + if index < self.size // 2: + current = self.head + for i in range(index): + current = current.next + else: + current = self.tail + for i in range(self.size - index - 1): + current = current.prev + + return current.val + + def addAtHead(self, val: int) -> None: + new_node = ListNode(val, None, self.head) + if self.head: + self.head.prev = new_node + else: + self.tail = new_node + self.head = new_node + self.size += 1 + + def addAtTail(self, val: int) -> None: + new_node = ListNode(val, self.tail, None) + if self.tail: + self.tail.next = new_node + else: + self.head = new_node + self.tail = new_node + self.size += 1 + + def addAtIndex(self, index: int, val: int) -> None: + if index < 0 or index > self.size: + return + + if index == 0: + self.addAtHead(val) + elif index == self.size: + self.addAtTail(val) + else: + if index < self.size // 2: + current = self.head + for i in range(index - 1): + current = current.next + else: + current = self.tail + for i in range(self.size - index): + current = current.prev + new_node = ListNode(val, current, current.next) + current.next.prev = new_node + current.next = new_node + self.size += 1 + + def deleteAtIndex(self, index: int) -> None: + if index < 0 or index >= self.size: + return + + if index == 0: + self.head = self.head.next + if self.head: + self.head.prev = None + else: + self.tail = None + elif index == self.size - 1: + self.tail = self.tail.prev + if self.tail: + self.tail.next = None + else: + self.head = None + else: + if index < self.size // 2: + current = self.head + for i in range(index): + current = current.next + else: + current = self.tail + for i in range(self.size - index - 1): + current = current.prev + current.prev.next = current.next + current.next.prev = current.prev + self.size -= 1 From 55ea26c5bdac70997c6ad9833448cdf3c1a7e0d1 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 17:18:48 -0500 Subject: [PATCH 065/154] =?UTF-8?q?Update=200206.=E7=BF=BB=E8=BD=AC?= =?UTF-8?q?=E9=93=BE=E8=A1=A8.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0206.翻转链表.md | 40 ++++++++++------------------------- 1 file changed, 11 insertions(+), 29 deletions(-) diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md index d558e783..0425e182 100644 --- a/problems/0206.翻转链表.md +++ b/problems/0206.翻转链表.md @@ -193,9 +193,9 @@ class Solution { } ``` -Python迭代法： +Python ```python -#双指针 +（版本一）双指针法 # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): @@ -205,7 +205,7 @@ class Solution: def reverseList(self, head: ListNode) -> ListNode: cur = head pre = None - while(cur!=None): + while cur: temp = cur.next # 保存一下 cur的下一个节点，因为接下来要改变cur->next cur.next = pre #反转 #更新pre、cur指针 @@ -217,6 +217,7 @@ class Solution: Python递归法： ```python +（版本二）递归法 # Definition for singly-linked list. # class ListNode: # def __init__(self, val=0, next=None): @@ -224,36 +225,17 @@ Python递归法： # self.next = next class Solution: def reverseList(self, head: ListNode) -> ListNode: - - def reverse(pre,cur): - if not cur: - return pre - - tmp = cur.next - cur.next = pre - - return reverse(cur,tmp) - - return reverse(None,head) + return self.reverse(head, None) + def reverse(self, cur: ListNode, pre: ListNode) -> ListNode: + if cur == None: + return pre + temp = cur.next + cur.next = pre + return self.reverse(temp, cur) ``` -Python递归法从后向前： -```python -# Definition for singly-linked list. -# class ListNode: -# def __init__(self, val=0, next=None): -# self.val = val -# self.next = next -class Solution: - def reverseList(self, head: Optional[ListNode]) -> Optional[ListNode]: - if not head or not head.next: return head - p = self.reverseList(head.next) - head.next.next = head - head.next = None - return p -``` Go： From d92104209fecd89626df38414d03c78371f251e5 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 17:20:34 -0500 Subject: [PATCH 066/154] =?UTF-8?q?Update=200024.=E4=B8=A4=E4=B8=A4?= =?UTF-8?q?=E4=BA=A4=E6=8D=A2=E9=93=BE=E8=A1=A8=E4=B8=AD=E7=9A=84=E8=8A=82?= =?UTF-8?q?=E7=82=B9.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0024.两两交换链表中的节点.md | 23 +++++++++---------- 1 file changed, 11 insertions(+), 12 deletions(-) diff --git a/problems/0024.两两交换链表中的节点.md b/problems/0024.两两交换链表中的节点.md index 4dc051aa..2c171dde 100644 --- a/problems/0024.两两交换链表中的节点.md +++ b/problems/0024.两两交换链表中的节点.md @@ -186,21 +186,20 @@ Python： class Solution: def swapPairs(self, head: ListNode) -> ListNode: - res = ListNode(next=head) - pre = res + dummy_head = ListNode(next=head) + current = dummy_head - # 必须有pre的下一个和下下个才能交换，否则说明已经交换结束了 - while pre.next and pre.next.next: - cur = pre.next - post = pre.next.next + # 必须有cur的下一个和下下个才能交换，否则说明已经交换结束了 + while current.next and current.next.next: + temp = current.next # 防止节点修改 + temp1 = current.next.next.next - # pre，cur，post对应最左，中间的，最右边的节点 - cur.next = post.next - post.next = cur - pre.next = post + current.next = current.next.next + current.next.next = temp + temp.next = temp1 + current = current.next.next + return dummy_head.next - pre = pre.next.next - return res.next ``` Go： From 190477400ebf0842343588c4e9e15c5eab412eb4 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 17:22:46 -0500 Subject: [PATCH 067/154] =?UTF-8?q?Update=200019.=E5=88=A0=E9=99=A4?= =?UTF-8?q?=E9=93=BE=E8=A1=A8=E7=9A=84=E5=80=92=E6=95=B0=E7=AC=ACN?= =?UTF-8?q?=E4=B8=AA=E8=8A=82=E7=82=B9.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...0019.删除链表的倒数第N个节点.md | 28 ++++++++++++------- 1 file changed, 18 insertions(+), 10 deletions(-) diff --git a/problems/0019.删除链表的倒数第N个节点.md b/problems/0019.删除链表的倒数第N个节点.md index a11ff8ba..c6f5bfc7 100644 --- a/problems/0019.删除链表的倒数第N个节点.md +++ b/problems/0019.删除链表的倒数第N个节点.md @@ -127,21 +127,29 @@ Python: # def __init__(self, val=0, next=None): # self.val = val # self.next = next + class Solution: def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode: - head_dummy = ListNode() - head_dummy.next = head - - slow, fast = head_dummy, head_dummy - while(n>=0): #fast先往前走n+1步 + # 创建一个虚拟节点，并将其下一个指针设置为链表的头部 + dummy_head = ListNode(0, head) + + # 创建两个指针，慢指针和快指针，并将它们初始化为虚拟节点 + slow = fast = dummy_head + + # 快指针比慢指针快 n+1 步 + for i in range(n+1): fast = fast.next - n -= 1 - while(fast!=None): + + # 移动两个指针，直到快速指针到达链表的末尾 + while fast: slow = slow.next fast = fast.next - #fast 走到结尾后，slow的下一个节点为倒数第N个节点 - slow.next = slow.next.next #删除 - return head_dummy.next + + # 通过更新第 (n-1) 个节点的 next 指针删除第 n 个节点 + slow.next = slow.next.next + + return dummy_head.next + ``` Go: ```Go From e4072d9a00b12c6de6949091378bdd8d95b5296c Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 17:25:36 -0500 Subject: [PATCH 068/154] =?UTF-8?q?Update=20=E9=9D=A2=E8=AF=95=E9=A2=9802.?= =?UTF-8?q?07.=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/面试题02.07.链表相交.md | 63 ++++++++++++++++++++++++- 1 file changed, 62 insertions(+), 1 deletion(-) diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index 30f5c467..4bcbd1f9 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -152,7 +152,7 @@ public class Solution { ### Python ```python - +（版本一）求长度，同时出发 class Solution: def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: lenA, lenB = 0, 0 @@ -178,7 +178,68 @@ class Solution: curB = curB.next return None ``` +```python +（版本二）求长度，同时出发 （代码复用） +class Solution: + def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: + lenA = self.getLength(headA) + lenB = self.getLength(headB) + + # 通过移动较长的链表，使两链表长度相等 + if lenA > lenB: + headA = self.moveForward(headA, lenA - lenB) + else: + headB = self.moveForward(headB, lenB - lenA) + + # 将两个头向前移动，直到它们相交 + while headA and headB: + if headA == headB: + return headA + headA = headA.next + headB = headB.next + + return None + + def getLength(self, head: ListNode) -> int: + length = 0 + while head: + length += 1 + head = head.next + return length + + def moveForward(self, head: ListNode, steps: int) -> ListNode: + while steps > 0: + head = head.next + steps -= 1 + return head +``` +```python +（版本三）等比例法 +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None +class Solution: + def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: + # 处理边缘情况 + if not headA or not headB: + return None + + # 在每个链表的头部初始化两个指针 + pointerA = headA + pointerB = headB + + # 遍历两个链表直到指针相交 + while pointerA != pointerB: + # 将指针向前移动一个节点 + pointerA = pointerA.next if pointerA else headB + pointerB = pointerB.next if pointerB else headA + + # 如果相交，指针将位于交点节点，如果没有交点，值为None + return pointerA +``` ### Go ```go From b2bfb8016642f388389250df57bad7e9cce82a3f Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 17:27:27 -0500 Subject: [PATCH 069/154] =?UTF-8?q?Update=200142.=E7=8E=AF=E5=BD=A2?= =?UTF-8?q?=E9=93=BE=E8=A1=A8II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0142.环形链表II.md | 50 ++++++++++++++++++++++++++------- 1 file changed, 40 insertions(+), 10 deletions(-) diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md index 46df4777..e80a715a 100644 --- a/problems/0142.环形链表II.md +++ b/problems/0142.环形链表II.md @@ -221,25 +221,55 @@ public class Solution { Python： ```python +（版本一）快慢指针法 +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + + class Solution: def detectCycle(self, head: ListNode) -> ListNode: - slow, fast = head, head + slow = head + fast = head + while fast and fast.next: slow = slow.next fast = fast.next.next - # 如果相遇 + + # If there is a cycle, the slow and fast pointers will eventually meet if slow == fast: - p = head - q = slow - while p!=q: - p = p.next - q = q.next - #你也可以return q - return p - + # Move one of the pointers back to the start of the list + slow = head + while slow != fast: + slow = slow.next + fast = fast.next + return slow + # If there is no cycle, return None return None ``` +```python +（版本二）集合法 +# Definition for singly-linked list. +# class ListNode: +# def __init__(self, x): +# self.val = x +# self.next = None + +class Solution: + def detectCycle(self, head: ListNode) -> ListNode: + visited = set() + + while head: + if head in visited: + return head + visited.add(head) + head = head.next + + return None +``` Go： ```go From dd58553088514af1a1a019fb9e6d4b2e41b0aeb5 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 19:26:39 -0500 Subject: [PATCH 070/154] =?UTF-8?q?Update=20=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E9=80=92=E5=BD=92=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树的递归遍历.md | 53 +++++++++++++--------------- 1 file changed, 24 insertions(+), 29 deletions(-) diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md index 5a9a670a..8d5a5985 100644 --- a/problems/二叉树的递归遍历.md +++ b/problems/二叉树的递归遍历.md @@ -174,50 +174,45 @@ class Solution { Python： ```python # 前序遍历-递归-LC144_二叉树的前序遍历 +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right + class Solution: def preorderTraversal(self, root: TreeNode) -> List[int]: - # 保存结果 - result = [] - - def traversal(root: TreeNode): - if root == None: - return - result.append(root.val) # 前序 - traversal(root.left) # 左 - traversal(root.right) # 右 + if not root: + return [] + + left = self.preorderTraversal(root.left) + right = self.preorderTraversal(root.right) + + return [root.val] + left + right - traversal(root) - return result # 中序遍历-递归-LC94_二叉树的中序遍历 class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: - result = [] + if root is None: + return [] - def traversal(root: TreeNode): - if root == None: - return - traversal(root.left) # 左 - result.append(root.val) # 中序 - traversal(root.right) # 右 + left = self.inorderTraversal(root.left) + right = self.inorderTraversal(root.right) - traversal(root) - return result + return left + [root.val] + right # 后序遍历-递归-LC145_二叉树的后序遍历 class Solution: def postorderTraversal(self, root: TreeNode) -> List[int]: - result = [] + if not root: + return [] - def traversal(root: TreeNode): - if root == None: - return - traversal(root.left) # 左 - traversal(root.right) # 右 - result.append(root.val) # 后序 + left = self.postorderTraversal(root.left) + right = self.postorderTraversal(root.right) - traversal(root) - return result + return left + right + [root.val] ``` Go： From dd1da7fc548c7ab7c1e0964f0ef50410be53e551 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 19:27:58 -0500 Subject: [PATCH 071/154] =?UTF-8?q?Update=20=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E9=80=92=E5=BD=92=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树的递归遍历.md | 6 +++++- 1 file changed, 5 insertions(+), 1 deletion(-) diff --git a/problems/二叉树的递归遍历.md b/problems/二叉树的递归遍历.md index 8d5a5985..92f342f0 100644 --- a/problems/二叉树的递归遍历.md +++ b/problems/二叉树的递归遍历.md @@ -191,7 +191,8 @@ class Solution: return [root.val] + left + right - +``` +```python # 中序遍历-递归-LC94_二叉树的中序遍历 class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: @@ -202,6 +203,9 @@ class Solution: right = self.inorderTraversal(root.right) return left + [root.val] + right +``` +```python + # 后序遍历-递归-LC145_二叉树的后序遍历 class Solution: From ce5b335b7258b7febb2dc831d68bfd3bb6161905 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 19:30:53 -0500 Subject: [PATCH 072/154] =?UTF-8?q?Update=20=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E8=BF=AD=E4=BB=A3=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/二叉树的迭代遍历.md | 8 ++++++-- 1 file changed, 6 insertions(+), 2 deletions(-) diff --git a/problems/二叉树的迭代遍历.md b/problems/二叉树的迭代遍历.md index 35cf4077..8b241465 100644 --- a/problems/二叉树的迭代遍历.md +++ b/problems/二叉树的迭代遍历.md @@ -258,7 +258,9 @@ class Solution: if node.left: stack.append(node.left) return result - +``` +```python + # 中序遍历-迭代-LC94_二叉树的中序遍历 class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: @@ -279,7 +281,9 @@ class Solution: # 取栈顶元素右结点 cur = cur.right return result - + ``` + ```python + # 后序遍历-迭代-LC145_二叉树的后序遍历 class Solution: def postorderTraversal(self, root: TreeNode) -> List[int]: From 0521f762d90b1af39fa4f134d219e2d208988e3b Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 19:48:57 -0500 Subject: [PATCH 073/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 3 +-- 1 file changed, 1 insertion(+), 2 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index e777c2a4..29deee11 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -178,12 +178,11 @@ python3代码： # self.val = val # self.left = left # self.right = right -from collections import deque class Solution: def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: if not root: return [] - queue = deque([root]) + queue = collections.deque([root]) result = [] while queue: level = [] From 42f85c8a8f8c637dbb484ca0d0c9ab88973c3a0b Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 19:56:26 -0500 Subject: [PATCH 074/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 36 ++++++++++++----------- 1 file changed, 19 insertions(+), 17 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 29deee11..a4164b2c 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -512,27 +512,29 @@ python代码： class Solution: """二叉树层序遍历II迭代解法""" +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: def levelOrderBottom(self, root: TreeNode) -> List[List[int]]: - results = [] if not root: - return results - - from collections import deque - que = deque([root]) - - while que: - result = [] - for _ in range(len(que)): - cur = que.popleft() - result.append(cur.val) + return [] + queue = collections.deque([root]) + result = [] + while queue: + level = [] + for _ in range(len(queue)): + cur = queue.popleft() + level.append(cur.val) if cur.left: - que.append(cur.left) + queue.append(cur.left) if cur.right: - que.append(cur.right) - results.append(result) - - results.reverse() - return results + queue.append(cur.right) + result.append(level) + return result[::-1] ``` Java： From 52092d0153828a14179b41d041e99121a9b4f1f2 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:01:07 -0500 Subject: [PATCH 075/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 46 +++++++++++------------ 1 file changed, 23 insertions(+), 23 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index a4164b2c..fdfde822 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -835,35 +835,35 @@ public: python代码： ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: def rightSideView(self, root: TreeNode) -> List[int]: if not root: return [] - - # deque来自collections模块，不在力扣平台时，需要手动写入 - # 'from collections import deque' 导入 - # deque相比list的好处是，list的pop(0)是O(n)复杂度，deque的popleft()是O(1)复杂度 - - quene = deque([root]) - out_list = [] - - while quene: - # 每次都取最后一个node就可以了 - node = quene[-1] - out_list.append(node.val) - - # 执行这个遍历的目的是获取下一层所有的node - for _ in range(len(quene)): - node = quene.popleft() + + queue = collections.deque([root]) + right_view = [] + + while queue: + level_size = len(queue) + + for i in range(level_size): + node = queue.popleft() + + if i == level_size - 1: + right_view.append(node.val) + if node.left: - quene.append(node.left) + queue.append(node.left) if node.right: - quene.append(node.right) - - return out_list - -# 执行用时：36 ms, 在所有 Python3 提交中击败了89.47%的用户 -# 内存消耗：14.6 MB, 在所有 Python3 提交中击败了96.65%的用户 + queue.append(node.right) + + return right_view ``` From 95616fa9a8a147f3255ead5523af0a4de51844b9 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:03:09 -0500 Subject: [PATCH 076/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 47 ++++++++++++++--------- 1 file changed, 29 insertions(+), 18 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index fdfde822..f4a5c466 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -1121,27 +1121,38 @@ python代码： class Solution: """二叉树层平均值迭代解法""" +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: def averageOfLevels(self, root: TreeNode) -> List[float]: - results = [] if not root: - return results + return [] - from collections import deque - que = deque([root]) - - while que: - size = len(que) - sum_ = 0 - for _ in range(size): - cur = que.popleft() - sum_ += cur.val - if cur.left: - que.append(cur.left) - if cur.right: - que.append(cur.right) - results.append(sum_ / size) - - return results + queue = collections.deque([root]) + averages = [] + + while queue: + size = len(queue) + level_sum = 0 + + for i in range(size): + node = queue.popleft() + + + level_sum += node.val + + if node.left: + queue.append(node.left) + if node.right: + queue.append(node.right) + + averages.append(level_sum / size) + + return averages ``` java: From 7be18e2dd372a17288c8cb215b52dc5a06c1a240 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:04:54 -0500 Subject: [PATCH 077/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 40 ++++++++++++++--------- 1 file changed, 24 insertions(+), 16 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index f4a5c466..ed2c683d 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -1426,28 +1426,36 @@ public: python代码： ```python +""" +# Definition for a Node. +class Node: + def __init__(self, val=None, children=None): + self.val = val + self.children = children +""" + class Solution: - """N叉树的层序遍历迭代法""" - def levelOrder(self, root: 'Node') -> List[List[int]]: - results = [] if not root: - return results + return [] - from collections import deque - que = deque([root]) + result = [] + queue = collections.deque([root]) - while que: - result = [] - for _ in range(len(que)): - cur = que.popleft() - result.append(cur.val) - # cur.children 是 Node 对象组成的列表，也可能为 None - if cur.children: - que.extend(cur.children) - results.append(result) + while queue: + level_size = len(queue) + level = [] - return results + for _ in range(level_size): + node = queue.popleft() + level.append(node.val) + + for child in node.children: + queue.append(child) + + result.append(level) + + return result ``` ```python From 994db141aa831866a5551d7c01c5d676252b9876 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:06:39 -0500 Subject: [PATCH 078/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 39 ++++++++++++++++------- 1 file changed, 27 insertions(+), 12 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index ed2c683d..6235ad21 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -1761,22 +1761,37 @@ public: python代码： ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: def largestValues(self, root: TreeNode) -> List[int]: - if root is None: + if not root: return [] - queue = [root] - out_list = [] + + result = [] + queue = collections.deque([root]) + while queue: - length = len(queue) - in_list = [] - for _ in range(length): - curnode = queue.pop(0) - in_list.append(curnode.val) - if curnode.left: queue.append(curnode.left) - if curnode.right: queue.append(curnode.right) - out_list.append(max(in_list)) - return out_list + level_size = len(queue) + max_val = float('-inf') + + for _ in range(level_size): + node = queue.popleft() + max_val = max(max_val, node.val) + + if node.left: + queue.append(node.left) + + if node.right: + queue.append(node.right) + + result.append(max_val) + + return result ``` java代码： From ffab57f2c330f35a624bd7c9ce7dc9d444f0ff79 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:09:12 -0500 Subject: [PATCH 079/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 48 ++++++++++++----------- 1 file changed, 26 insertions(+), 22 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 6235ad21..47969b25 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -2096,36 +2096,40 @@ class Solution { python代码： ```python -# 层序遍历解法 +""" +# Definition for a Node. +class Node: + def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None): + self.val = val + self.left = left + self.right = right + self.next = next +""" class Solution: def connect(self, root: 'Node') -> 'Node': if not root: - return None - queue = [root] + return root + + queue = collections.deque([root]) + while queue: - n = len(queue) - for i in range(n): - node = queue.pop(0) + level_size = len(queue) + prev = None + + for i in range(level_size): + node = queue.popleft() + + if prev: + prev.next = node + + prev = node + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) - if i == n - 1: - break - node.next = queue[0] - return root - -# 链表解法 -class Solution: - def connect(self, root: 'Node') -> 'Node': - first = root - while first: - cur = first - while cur: # 遍历每一层的节点 - if cur.left: cur.left.next = cur.right # 找左节点的next - if cur.right and cur.next: cur.right.next = cur.next.left # 找右节点的next - cur = cur.next # cur同层移动到下一节点 - first = first.left # 从本层扩展到下一层 + return root ``` From 65085c4e9bcf8c478fe67caad7dd0dceaa5911e0 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:10:51 -0500 Subject: [PATCH 080/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 44 ++++++++++++++++------- 1 file changed, 32 insertions(+), 12 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 47969b25..89e7a126 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -2381,21 +2381,41 @@ python代码： ```python # 层序遍历解法 +""" +# Definition for a Node. +class Node: + def __init__(self, val: int = 0, left: 'Node' = None, right: 'Node' = None, next: 'Node' = None): + self.val = val + self.left = left + self.right = right + self.next = next +""" + class Solution: def connect(self, root: 'Node') -> 'Node': if not root: - return None - queue = [root] - while queue: # 遍历每一层 - length = len(queue) - tail = None # 每一层维护一个尾节点 - for i in range(length): # 遍历当前层 - curnode = queue.pop(0) - if tail: - tail.next = curnode # 让尾节点指向当前节点 - tail = curnode # 让当前节点成为尾节点 - if curnode.left : queue.append(curnode.left) - if curnode.right: queue.append(curnode.right) + return root + + queue = collections.deque([root]) + + while queue: + level_size = len(queue) + prev = None + + for i in range(level_size): + node = queue.popleft() + + if prev: + prev.next = node + + prev = node + + if node.left: + queue.append(node.left) + + if node.right: + queue.append(node.right) + return root ``` From 61173277185bf7948ffb576212cc31fd383778e3 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:12:16 -0500 Subject: [PATCH 081/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 31 ++++++++++++++--------- 1 file changed, 19 insertions(+), 12 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 89e7a126..13694207 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -2664,24 +2664,31 @@ class Solution { Python： ```python 3 +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: def maxDepth(self, root: TreeNode) -> int: - if root == None: + if not root: return 0 - - queue_ = [root] + depth = 0 - while queue_: - length = len(queue_) - for i in range(length): - cur = queue_.pop(0) - sub.append(cur.val) - #子节点入队列 - if cur.left: queue_.append(cur.left) - if cur.right: queue_.append(cur.right) + queue = collections.deque([root]) + + while queue: depth += 1 - + for _ in range(len(queue)): + node = queue.popleft() + if node.left: + queue.append(node.left) + if node.right: + queue.append(node.right) + return depth + ``` Go： From de7c67c35a5d3d0e423072d0b7cb02876c628786 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:13:38 -0500 Subject: [PATCH 082/154] =?UTF-8?q?Update=200102.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E5=B1=82=E5=BA=8F=E9=81=8D=E5=8E=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0102.二叉树的层序遍历.md | 33 ++++++++++++----------- 1 file changed, 18 insertions(+), 15 deletions(-) diff --git a/problems/0102.二叉树的层序遍历.md b/problems/0102.二叉树的层序遍历.md index 13694207..c2ad9508 100644 --- a/problems/0102.二叉树的层序遍历.md +++ b/problems/0102.二叉树的层序遍历.md @@ -2938,23 +2938,26 @@ Python 3： # self.right = right class Solution: def minDepth(self, root: TreeNode) -> int: - if root == None: + if not root: return 0 + depth = 0 + queue = collections.deque([root]) + + while queue: + depth += 1 + for _ in range(len(queue)): + node = queue.popleft() + + if not node.left and not node.right: + return depth + + if node.left: + queue.append(node.left) + + if node.right: + queue.append(node.right) - #根节点的深度为1 - queue_ = [(root,1)] - while queue_: - cur, depth = queue_.pop(0) - - if cur.left == None and cur.right == None: - return depth - #先左子节点，由于左子节点没有孩子，则就是这一层了 - if cur.left: - queue_.append((cur.left,depth + 1)) - if cur.right: - queue_.append((cur.right,depth + 1)) - - return 0 + return depth ``` Go： From 54bcab13e37ee1e609b567411b968bdb5a86abac Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:25:30 -0500 Subject: [PATCH 083/154] =?UTF-8?q?Update=200226.=E7=BF=BB=E8=BD=AC?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0226.翻转二叉树.md | 163 +++++++++++++++++++++++-------- 1 file changed, 120 insertions(+), 43 deletions(-) diff --git a/problems/0226.翻转二叉树.md b/problems/0226.翻转二叉树.md index 16a5be57..63baa409 100644 --- a/problems/0226.翻转二叉树.md +++ b/problems/0226.翻转二叉树.md @@ -314,81 +314,158 @@ class Solution { 递归法：前序遍历： ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: def invertTree(self, root: TreeNode) -> TreeNode: if not root: return None - root.left, root.right = root.right, root.left #中 - self.invertTree(root.left) #左 - self.invertTree(root.right) #右 + root.left, root.right = root.right, root.left + self.invertTree(root.left) + self.invertTree(root.right) return root ``` -递归法：后序遍历： +迭代法：前序遍历： ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: def invertTree(self, root: TreeNode) -> TreeNode: - if root is None: + if not root: + return None + stack = [root] + while stack: + node = stack.pop() + node.left, node.right = node.right, node.left + if node.left: + stack.append(node.left) + if node.right: + stack.append(node.right) + return root +``` + + +递归法：中序遍历： +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def invertTree(self, root: TreeNode) -> TreeNode: + if not root: + return None + self.invertTree(root.left) + root.left, root.right = root.right, root.left + self.invertTree(root.left) + return root +``` + +迭代法：中序遍历： +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def invertTree(self, root: TreeNode) -> TreeNode: + if not root: + return None + stack = [root] + while stack: + node = stack.pop() + if node.left: + stack.append(node.left) + node.left, node.right = node.right, node.left + if node.left: + stack.append(node.left) + return root +``` + + +递归法：后序遍历： +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def invertTree(self, root: TreeNode) -> TreeNode: + if not root: return None self.invertTree(root.left) self.invertTree(root.right) root.left, root.right = root.right, root.left - return root + return root ``` -迭代法：深度优先遍历（前序遍历）： +迭代法：后序遍历： ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: def invertTree(self, root: TreeNode) -> TreeNode: if not root: - return root - st = [] - st.append(root) - while st: - node = st.pop() - node.left, node.right = node.right, node.left #中 - if node.right: - st.append(node.right) #右 + return None + stack = [root] + while stack: + node = stack.pop() if node.left: - st.append(node.left) #左 + stack.append(node.left) + if node.right: + stack.append(node.right) + node.left, node.right = node.right, node.left + return root ``` + + + + 迭代法：广度优先遍历（层序遍历）： ```python -import collections +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: def invertTree(self, root: TreeNode) -> TreeNode: - queue = collections.deque() #使用deque() - if root: - queue.append(root) + if not root: + return None + + queue = collections.deque([root]) while queue: - size = len(queue) - for i in range(size): + for i in range(len(queue)): node = queue.popleft() - node.left, node.right = node.right, node.left #节点处理 - if node.left: - queue.append(node.left) - if node.right: - queue.append(node.right) - return root -``` -迭代法：广度优先遍历（层序遍历），和之前的层序遍历写法一致： -```python -class Solution: - def invertTree(self, root: Optional[TreeNode]) -> Optional[TreeNode]: - if not root: return root - from collections import deque - que=deque([root]) - while que: - size=len(que) - for i in range(size): - cur=que.popleft() - cur.left, cur.right = cur.right, cur.left - if cur.left: que.append(cur.left) - if cur.right: que.append(cur.right) + node.left, node.right = node.right, node.left + if node.left: queue.append(node.left) + if node.right: queue.append(node.right) return root + ``` + ### Go 递归版本的前序遍历 From 918d0ec0faf27870212b11c4820d4ae956ba8da3 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 20:40:43 -0500 Subject: [PATCH 084/154] =?UTF-8?q?Update=200101.=E5=AF=B9=E7=A7=B0?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0101.对称二叉树.md | 40 ++++++++++++++++++-------------- 1 file changed, 23 insertions(+), 17 deletions(-) diff --git a/problems/0101.对称二叉树.md b/problems/0101.对称二叉树.md index b75e9ff2..81ca79a2 100644 --- a/problems/0101.对称二叉树.md +++ b/problems/0101.对称二叉树.md @@ -442,25 +442,31 @@ class Solution: 层次遍历 ```python class Solution: - def isSymmetric(self, root: Optional[TreeNode]) -> bool: + def isSymmetric(self, root: TreeNode) -> bool: if not root: return True - - que = [root] - while que: - this_level_length = len(que) - for i in range(this_level_length // 2): - # 要么其中一个是None但另外一个不是 - if (not que[i] and que[this_level_length - 1 - i]) or (que[i] and not que[this_level_length - 1 - i]): - return False - # 要么两个都不是None - if que[i] and que[i].val != que[this_level_length - 1 - i].val: - return False - for i in range(this_level_length): - if not que[i]: continue - que.append(que[i].left) - que.append(que[i].right) - que = que[this_level_length:] + + queue = collections.deque([root.left, root.right]) + + while queue: + level_size = len(queue) + + if level_size % 2 != 0: + return False + + level_vals = [] + for i in range(level_size): + node = queue.popleft() + if node: + level_vals.append(node.val) + queue.append(node.left) + queue.append(node.right) + else: + level_vals.append(None) + + if level_vals != level_vals[::-1]: + return False + return True ``` From aac7378cb62ed2ec36fc1012f0057844297bda67 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 21:40:49 -0500 Subject: [PATCH 085/154] =?UTF-8?q?Update=200104.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E5=A4=A7=E6=B7=B1=E5=BA=A6.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0104.二叉树的最大深度.md | 25 +++++++++++++++-------- 1 file changed, 16 insertions(+), 9 deletions(-) diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md index 36578fd3..2294c1d9 100644 --- a/problems/0104.二叉树的最大深度.md +++ b/problems/0104.二叉树的最大深度.md @@ -419,26 +419,33 @@ class solution: return 1 + max(self.maxdepth(root.left), self.maxdepth(root.right)) ``` -迭代法： +层序遍历迭代法： ```python -import collections -class solution: - def maxdepth(self, root: treenode) -> int: +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def maxDepth(self, root: TreeNode) -> int: if not root: return 0 - depth = 0 #记录深度 - queue = collections.deque() - queue.append(root) + + depth = 0 + queue = collections.deque([root]) + while queue: - size = len(queue) depth += 1 - for i in range(size): + for _ in range(len(queue)): node = queue.popleft() if node.left: queue.append(node.left) if node.right: queue.append(node.right) + return depth + ``` ### 559.n叉树的最大深度 From 5da51519b5692ef2f450ba38b6299452a00f3b22 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 21:48:04 -0500 Subject: [PATCH 086/154] =?UTF-8?q?Update=200104.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E5=A4=A7=E6=B7=B1=E5=BA=A6.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0104.二叉树的最大深度.md | 15 +++++++++------ 1 file changed, 9 insertions(+), 6 deletions(-) diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md index 2294c1d9..c4d94d4d 100644 --- a/problems/0104.二叉树的最大深度.md +++ b/problems/0104.二叉树的最大深度.md @@ -452,14 +452,17 @@ class Solution: 递归法： ```python -class solution: - def maxdepth(self, root: 'node') -> int: +class Solution: + def maxDepth(self, root: 'Node') -> int: if not root: return 0 - depth = 0 - for i in range(len(root.children)): - depth = max(depth, self.maxdepth(root.children[i])) - return depth + 1 + + max_depth = 1 + + for child in root.children: + max_depth = max(max_depth, self.maxDepth(child) + 1) + + return max_depth ``` 迭代法： From be27cc547daf4c82a048b71f201c93ed3b2b3e2e Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 21:57:13 -0500 Subject: [PATCH 087/154] =?UTF-8?q?Update=200104.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E5=A4=A7=E6=B7=B1=E5=BA=A6.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0104.二叉树的最大深度.md | 33 ++++++++++++++--------- 1 file changed, 21 insertions(+), 12 deletions(-) diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md index c4d94d4d..c55ddb3f 100644 --- a/problems/0104.二叉树的最大深度.md +++ b/problems/0104.二叉树的最大深度.md @@ -467,22 +467,31 @@ class Solution: 迭代法： ```python -import collections -class solution: - def maxdepth(self, root: 'node') -> int: - queue = collections.deque() - if root: - queue.append(root) - depth = 0 #记录深度 +""" +# Definition for a Node. +class Node: + def __init__(self, val=None, children=None): + self.val = val + self.children = children +""" + +class Solution: + def maxDepth(self, root: TreeNode) -> int: + if not root: + return 0 + + depth = 0 + queue = collections.deque([root]) + while queue: - size = len(queue) depth += 1 - for i in range(size): + for _ in range(len(queue)): node = queue.popleft() - for j in range(len(node.children)): - if node.children[j]: - queue.append(node.children[j]) + for child in node.children: + queue.append(child) + return depth + ``` 使用栈来模拟后序遍历依然可以 From 1b1b51750db1afcb812ba1250ea4fdc3e0395425 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 3 May 2023 22:04:25 -0500 Subject: [PATCH 088/154] =?UTF-8?q?Update=200104.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E5=A4=A7=E6=B7=B1=E5=BA=A6.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0104.二叉树的最大深度.md | 48 ++++++++++++----------- 1 file changed, 25 insertions(+), 23 deletions(-) diff --git a/problems/0104.二叉树的最大深度.md b/problems/0104.二叉树的最大深度.md index c55ddb3f..b6208169 100644 --- a/problems/0104.二叉树的最大深度.md +++ b/problems/0104.二叉树的最大深度.md @@ -494,30 +494,32 @@ class Solution: ``` -使用栈来模拟后序遍历依然可以 +使用栈 ```python -class solution: - def maxdepth(self, root: 'node') -> int: - st = [] - if root: - st.append(root) - depth = 0 - result = 0 - while st: - node = st.pop() - if node != none: - st.append(node) #中 - st.append(none) - depth += 1 - for i in range(len(node.children)): #处理孩子 - if node.children[i]: - st.append(node.children[i]) - - else: - node = st.pop() - depth -= 1 - result = max(result, depth) - return result +""" +# Definition for a Node. +class Node: + def __init__(self, val=None, children=None): + self.val = val + self.children = children +""" + +class Solution: + def maxDepth(self, root: 'Node') -> int: + if not root: + return 0 + + max_depth = 0 + + stack = [(root, 1)] + + while stack: + node, depth = stack.pop() + max_depth = max(max_depth, depth) + for child in node.children: + stack.append((child, depth + 1)) + + return max_depth ``` From 75d137eb1c0464de0321f611ed66125e322936e7 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 00:23:53 -0500 Subject: [PATCH 089/154] =?UTF-8?q?Update=200111.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E5=B0=8F=E6=B7=B1=E5=BA=A6.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0111.二叉树的最小深度.md | 59 +++++++++++++++-------- 1 file changed, 39 insertions(+), 20 deletions(-) diff --git a/problems/0111.二叉树的最小深度.md b/problems/0111.二叉树的最小深度.md index de36c6f2..bda12ff0 100644 --- a/problems/0111.二叉树的最小深度.md +++ b/problems/0111.二叉树的最小深度.md @@ -300,44 +300,63 @@ class Solution { 递归法： ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: def minDepth(self, root: TreeNode) -> int: if not root: return 0 + if not root.left and not root.right: return 1 - - min_depth = 10**9 + + left_depth = float('inf') + right_depth = float('inf') + if root.left: - min_depth = min(self.minDepth(root.left), min_depth) # 获得左子树的最小高度 + left_depth = self.minDepth(root.left) if root.right: - min_depth = min(self.minDepth(root.right), min_depth) # 获得右子树的最小高度 - return min_depth + 1 + right_depth = self.minDepth(root.right) + + return 1 + min(left_depth, right_depth) + ``` 迭代法： ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: def minDepth(self, root: TreeNode) -> int: if not root: return 0 - que = deque() - que.append(root) - res = 1 - - while que: - for _ in range(len(que)): - node = que.popleft() - # 当左右孩子都为空的时候，说明是最低点的一层了，退出 + depth = 0 + queue = collections.deque([root]) + + while queue: + depth += 1 + for _ in range(len(queue)): + node = queue.popleft() + if not node.left and not node.right: - return res - if node.left is not None: - que.append(node.left) - if node.right is not None: - que.append(node.right) - res += 1 - return res + return depth + + if node.left: + queue.append(node.left) + + if node.right: + queue.append(node.right) + + return depth ``` From d15c54f43fe0df608d207f269e8bb589c5cebd3d Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 00:26:20 -0500 Subject: [PATCH 090/154] =?UTF-8?q?Update=200111.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E5=B0=8F=E6=B7=B1=E5=BA=A6.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0111.二叉树的最小深度.md | 33 +++++++++++++++++++++++ 1 file changed, 33 insertions(+) diff --git a/problems/0111.二叉树的最小深度.md b/problems/0111.二叉树的最小深度.md index bda12ff0..0c086f1b 100644 --- a/problems/0111.二叉树的最小深度.md +++ b/problems/0111.二叉树的最小深度.md @@ -359,6 +359,39 @@ class Solution: return depth ``` +迭代法： + +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right + +class Solution: + def minDepth(self, root: TreeNode) -> int: + if not root: + return 0 + + queue = collections.deque([(root, 1)]) + + while queue: + node, depth = queue.popleft() + + # Check if the node is a leaf node + if not node.left and not node.right: + return depth + + # Add left and right child to the queue + if node.left: + queue.append((node.left, depth+1)) + if node.right: + queue.append((node.right, depth+1)) + + return 0 + +``` ## Go From 3d59f732e0479c12657f5a6020b4ed5784222780 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 01:35:13 -0500 Subject: [PATCH 091/154] =?UTF-8?q?Update=200222.=E5=AE=8C=E5=85=A8?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E7=9A=84=E8=8A=82=E7=82=B9=E4=B8=AA?= =?UTF-8?q?=E6=95=B0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0222.完全二叉树的节点个数.md | 14 ++++++++++++++ 1 file changed, 14 insertions(+) diff --git a/problems/0222.完全二叉树的节点个数.md b/problems/0222.完全二叉树的节点个数.md index d89a9bce..795a6f37 100644 --- a/problems/0222.完全二叉树的节点个数.md +++ b/problems/0222.完全二叉树的节点个数.md @@ -393,6 +393,20 @@ class Solution: # 利用完全二叉树特性 return 2**count-1 return 1+self.countNodes(root.left)+self.countNodes(root.right) ``` +完全二叉树写法3 +```python +class Solution: # 利用完全二叉树特性 + def countNodes(self, root: TreeNode) -> int: + if not root: return 0 + count = 0 + left = root.left; right = root.right + while left and right: + count+=1 + left = left.left; right = right.right + if not left and not right: # 如果同时到底说明是满二叉树，反之则不是 + return (2< Date: Thu, 4 May 2023 21:49:12 -0500 Subject: [PATCH 092/154] =?UTF-8?q?Update=200110.=E5=B9=B3=E8=A1=A1?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index 804c95eb..a3bc77fb 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -532,6 +532,24 @@ class Solution: else: return 1 + max(left_height, right_height) ``` +递归法精简版： + +```python +class Solution: + def isBalanced(self, root: TreeNode) -> bool: + return self.height(root) != -1 + def height(self, node: TreeNode) -> int: + if not node: + return 0 + left = self.height(node.left) + if left == -1: + return -1 + right = self.height(node.right) + if right == -1 or abs(left - right) > 1: + return -1 + return max(left, right) + 1 +``` + 迭代法： From debaa5e4a0cb16a58f226e04b0865ce1b423f122 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 22:54:49 -0500 Subject: [PATCH 093/154] =?UTF-8?q?Update=200257.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=89=80=E6=9C=89=E8=B7=AF=E5=BE=84.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0257.二叉树的所有路径.md | 65 +++++++++++++++++++++++ 1 file changed, 65 insertions(+) diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md index 8c542cea..c396f4a0 100644 --- a/problems/0257.二叉树的所有路径.md +++ b/problems/0257.二叉树的所有路径.md @@ -468,6 +468,71 @@ class Solution { ``` --- ## Python: + + +递归法+回溯（版本一） +```Python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +import copy +from typing import List, Optional + +class Solution: + def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]: + if not root: + return [] + result = [] + self.generate_paths(root, [], result) + return result + + def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None: + path.append(node.val) + if not node.left and not node.right: + result.append('->'.join(map(str, path))) + if node.left: + self.generate_paths(node.left, copy.copy(path), result) + if node.right: + self.generate_paths(node.right, copy.copy(path), result) + path.pop() + + +``` +递归法+回溯（版本二） +```Python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +import copy +from typing import List, Optional + +class Solution: + def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]: + if not root: + return [] + result = [] + self.generate_paths(root, [], result) + return result + + def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None: + if not node: + return + path.append(node.val) + if not node.left and not node.right: + result.append('->'.join(map(str, path))) + else: + self.generate_paths(node.left, copy.copy(path), result) + self.generate_paths(node.right, copy.copy(path), result) + path.pop() + +``` + 递归法+隐形回溯 ```Python # Definition for a binary tree node. From 3b0000a280b2e6a5b1bb033556cbc77e793736fb Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Thu, 4 May 2023 23:26:06 -0500 Subject: [PATCH 094/154] =?UTF-8?q?Update=200404.=E5=B7=A6=E5=8F=B6?= =?UTF-8?q?=E5=AD=90=E4=B9=8B=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0404.左叶子之和.md | 26 +++++++++++++++++--------- 1 file changed, 17 insertions(+), 9 deletions(-) diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md index cf5441c4..617978b7 100644 --- a/problems/0404.左叶子之和.md +++ b/problems/0404.左叶子之和.md @@ -250,19 +250,27 @@ class Solution { **递归后序遍历** ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: - def sumOfLeftLeaves(self, root: TreeNode) -> int: - if not root: + def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int: + if not root: return 0 - left_left_leaves_sum = self.sumOfLeftLeaves(root.left) # 左 - right_left_leaves_sum = self.sumOfLeftLeaves(root.right) # 右 - - cur_left_leaf_val = 0 - if root.left and not root.left.left and not root.left.right: - cur_left_leaf_val = root.left.val + # 检查根节点的左子节点是否为叶节点 + if root.left and not root.left.left and not root.left.right: + left_val = root.left.val + else: + left_val = self.sumOfLeftLeaves(root.left) - return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum # 中 + # 递归地计算右子树左叶节点的和 + right_val = self.sumOfLeftLeaves(root.right) + + return left_val + right_val ``` **迭代** From 4224c17e54e164141ca2f6a08978402ae115fea5 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 19:18:26 -0500 Subject: [PATCH 095/154] =?UTF-8?q?Update=20=E9=9D=A2=E8=AF=95=E9=A2=9802.?= =?UTF-8?q?07.=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/面试题02.07.链表相交.md | 36 ++++++++++++++++++++++++- 1 file changed, 35 insertions(+), 1 deletion(-) diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index 4bcbd1f9..dda7f2ad 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -214,7 +214,41 @@ class Solution: return head ``` ```python -（版本三）等比例法 +（版本三）求长度，同时出发 （代码复用 + 精简） +class Solution: + def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: + dis = self.getLength(headA) - self.getLength(headB) + + # 通过移动较长的链表，使两链表长度相等 + if dis > 0: + headA = self.moveForward(headA, dis) + else: + headB = self.moveForward(headB, abs(dis)) + + # 将两个头向前移动，直到它们相交 + while headA and headB: + if headA == headB: + return headA + headA = headA.next + headB = headB.next + + return None + + def getLength(self, head: ListNode) -> int: + length = 0 + while head: + length += 1 + head = head.next + return length + + def moveForward(self, head: ListNode, steps: int) -> ListNode: + while steps > 0: + head = head.next + steps -= 1 + return head +``` +```python +（版本四）等比例法 # Definition for singly-linked list. # class ListNode: # def __init__(self, x): From d590d268d717b0241c5d900f70e9356bf3c88903 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 19:50:20 -0500 Subject: [PATCH 096/154] =?UTF-8?q?Update=200242.=E6=9C=89=E6=95=88?= =?UTF-8?q?=E7=9A=84=E5=AD=97=E6=AF=8D=E5=BC=82=E4=BD=8D=E8=AF=8D.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0242.有效的字母异位词.md | 55 +++++++++++++++++++---- 1 file changed, 47 insertions(+), 8 deletions(-) diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md index 1006ea35..0e5683c7 100644 --- a/problems/0242.有效的字母异位词.md +++ b/problems/0242.有效的字母异位词.md @@ -122,6 +122,21 @@ class Solution { ``` Python： +(版本一) 使用数组 +```python +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + record = [0] * 26 + for i in range(len(s)): + record[ord(s[i]) - ord("a")] += 1 + for i in range(len(t)): + record[ord(t[i]) - ord("a")] -= 1 + for i in range(26): + if record[i] != 0: + return False + return True +``` +(版本二) 使用数组 ```python class Solution: def isAnagram(self, s: str, t: str) -> bool: @@ -138,12 +153,13 @@ class Solution: return True ``` -Python写法二（没有使用数组作为哈希表，只是介绍defaultdict这样一种解题思路）： +(版本三) 使用defaultdict ```python +from collections import defaultdict + class Solution: def isAnagram(self, s: str, t: str) -> bool: - from collections import defaultdict s_dict = defaultdict(int) t_dict = defaultdict(int) @@ -156,17 +172,40 @@ class Solution: return s_dict == t_dict ``` -Python写法三(没有使用数组作为哈希表，只是介绍Counter这种更方便的解题思路)： +(版本四) 使用字典 ```python -class Solution(object): +class Solution: def isAnagram(self, s: str, t: str) -> bool: - from collections import Counter - a_count = Counter(s) - b_count = Counter(t) - return a_count == b_count + if len(s) != len(t): + return False + + hash_table_s = {} + hash_table_t = {} + + for i in range(len(s)): + hash_table_s[s[i]] = hash_table_s.get(s[i], 0) + 1 + hash_table_t[t[i]] = hash_table_t.get(t[i], 0) + 1 + + return hash_table_s == hash_table_t ``` +(版本五) 使用排序 + +```python +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + return sorted(s) == sorted(t) +``` +(版本六) 使用Counter + +```python +from collections import Counter + +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + return Counter(s) == Counter(t) +``` Go： ```go From 3065cf38604dd672ad128a445b98c356604eb24e Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 20:04:47 -0500 Subject: [PATCH 097/154] =?UTF-8?q?Update=200349.=E4=B8=A4=E4=B8=AA?= =?UTF-8?q?=E6=95=B0=E7=BB=84=E7=9A=84=E4=BA=A4=E9=9B=86.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0349.两个数组的交集.md | 34 ++++++++++++++++++-------- 1 file changed, 24 insertions(+), 10 deletions(-) diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md index 0da0e30c..c2f6ef46 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -160,22 +160,29 @@ class Solution { ``` Python3： +（版本一） 使用字典和集合 ```python class Solution: def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: - val_dict = {} - ans = [] + # 使用哈希表存储一个数组中的所有元素 + table = {} for num in nums1: - val_dict[num] = 1 - - for num in nums2: - if num in val_dict.keys() and val_dict[num] == 1: - ans.append(num) - val_dict[num] = 0 + table[num] = table.get(num, 0) + 1 - return ans + # 使用集合存储结果 + res = set() + for num in nums2: + if num in table: + res.add(num) + del table[num] + + return list(res) +``` +（版本二） 使用数组 + +```python -class Solution: # 使用数组方法 +class Solution: def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: count1 = [0]*1001 count2 = [0]*1001 @@ -190,7 +197,14 @@ class Solution: # 使用数组方法 return result ``` +（版本三） 使用集合 +```python +class Solution: + def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: + return list(set(nums1) & set(nums2)) + +``` Go： ```go From eeae3282a87c334f726fc3c8cd5dabd5b71c4612 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 20:25:46 -0500 Subject: [PATCH 098/154] =?UTF-8?q?Update=200202.=E5=BF=AB=E4=B9=90?= =?UTF-8?q?=E6=95=B0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0202.快乐数.md | 96 ++++++++++++++++++++++++++++++-------- 1 file changed, 76 insertions(+), 20 deletions(-) diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md index 5ac29e86..7fe8cd8d 100644 --- a/problems/0202.快乐数.md +++ b/problems/0202.快乐数.md @@ -108,23 +108,14 @@ class Solution { ``` Python： +(版本一)使用集合 ```python class Solution: - def isHappy(self, n: int) -> bool: - def calculate_happy(num): - sum_ = 0 - - # 从个位开始依次取，平方求和 - while num: - sum_ += (num % 10) ** 2 - num = num // 10 - return sum_ - - # 记录中间结果 + def isHappy(self, n: int) -> bool: record = set() while True: - n = calculate_happy(n) + n = self.get_sum(n) if n == 1: return True @@ -134,21 +125,86 @@ class Solution: else: record.add(n) -# python的另一种写法 - 通过字符串来计算各位平方和 + def get_sum(self,n: int) -> int: + new_num = 0 + while n: + n, r = divmod(n, 10) + new_num += r ** 2 + return new_num + ``` + (版本二)使用集合 + ```python +class Solution: + def isHappy(self, n: int) -> bool: + record = set() + while n not in record: + record.add(n) + new_num = 0 + n_str = str(n) + for i in n_str: + new_num+=int(i)**2 + if new_num==1: return True + else: n = new_num + return False +``` + (版本三)使用数组 + ```python class Solution: def isHappy(self, n: int) -> bool: record = [] while n not in record: record.append(n) - newn = 0 - nn = str(n) - for i in nn: - newn+=int(i)**2 - if newn==1: return True - n = newn + new_num = 0 + n_str = str(n) + for i in n_str: + new_num+=int(i)**2 + if new_num==1: return True + else: n = new_num return False ``` - + (版本四)使用快慢指针 + ```python +class Solution: + def isHappy(self, n: int) -> bool: + slow = n + fast = n + while self.get_sum(fast) != 1 and self.get_sum(self.get_sum(fast)): + slow = self.get_sum(slow) + fast = self.get_sum(self.get_sum(fast)) + if slow == fast: + return False + return True + def get_sum(self,n: int) -> int: + new_num = 0 + while n: + n, r = divmod(n, 10) + new_num += r ** 2 + return new_num +``` + (版本五)使用集合+精简 + ```python +class Solution: + def isHappy(self, n: int) -> bool: + seen = set() + while n != 1: + n = sum(int(i) ** 2 for i in str(n)) + if n in seen: + return False + seen.add(n) + return True +``` + (版本六)使用数组+精简 + ```python +class Solution: + def isHappy(self, n: int) -> bool: + seen = [] + while n != 1: + n = sum(int(i) ** 2 for i in str(n)) + if n in seen: + return False + seen.append(n) + return True +``` Go： ```go func isHappy(n int) bool { From 7118f36b66e1def134fda97e992affbd0c5cd79b Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 20:38:05 -0500 Subject: [PATCH 099/154] =?UTF-8?q?Update=200001.=E4=B8=A4=E6=95=B0?= =?UTF-8?q?=E4=B9=8B=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0001.两数之和.md | 49 ++++++++++++++++++++++++++++++++++- 1 file changed, 48 insertions(+), 1 deletion(-) diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md index eea3ba7a..1af8787b 100644 --- a/problems/0001.两数之和.md +++ b/problems/0001.两数之和.md @@ -151,7 +151,7 @@ public int[] twoSum(int[] nums, int target) { ``` Python： - +（版本一） 使用字典 ```python class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: @@ -163,6 +163,53 @@ class Solution: records[value] = index # 遍历当前元素，并在map中寻找是否有匹配的key return [] ``` +（版本二）使用集合 +```python +class Solution: + def twoSum(self, nums: List[int], target: int) -> List[int]: + #创建一个集合来存储我们目前看到的数字 + seen = set() + for i, num in enumerate(nums): + complement = target - num + if complement in seen: + return [nums.index(complement), i] + seen.add(num) +``` +（版本三）使用双指针 +```python +class Solution: + def twoSum(self, nums: List[int], target: int) -> List[int]: + # 对输入列表进行排序 + nums_sorted = sorted(nums) + + # 使用双指针 + left = 0 + right = len(nums_sorted) - 1 + while left < right: + current_sum = nums_sorted[left] + nums_sorted[right] + if current_sum == target: + # 如果和等于目标数，则返回两个数的下标 + left_index = nums.index(nums_sorted[left]) + right_index = nums.index(nums_sorted[right]) + if left_index == right_index: + right_index = nums[left_index+1:].index(nums_sorted[right]) + left_index + 1 + return [left_index, right_index] + elif current_sum < target: + # 如果总和小于目标，则将左侧指针向右移动 + left += 1 + else: + # 如果总和大于目标值，则将右指针向左移动 + right -= 1 +``` +（版本四）暴力法 +```python +class Solution: + def twoSum(self, nums: List[int], target: int) -> List[int]: + for i in range(len(nums)): + for j in range(i+1, len(nums)): + if nums[i] + nums[j] == target: + return [i,j] +``` Go： From 0905d2c7871daca4adb18566f86216a83e7894bc Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 20:46:57 -0500 Subject: [PATCH 100/154] =?UTF-8?q?Update=200454.=E5=9B=9B=E6=95=B0?= =?UTF-8?q?=E7=9B=B8=E5=8A=A0II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0454.四数相加II.md | 14 ++++++-------- 1 file changed, 6 insertions(+), 8 deletions(-) diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index 411b60e8..7818c02b 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -125,11 +125,11 @@ class Solution { ``` Python： - +（版本一） 使用字典 ```python class Solution(object): def fourSumCount(self, nums1, nums2, nums3, nums4): - # use a dict to store the elements in nums1 and nums2 and their sum + # 使用字典存储nums1和nums2中的元素及其和 hashmap = dict() for n1 in nums1: for n2 in nums2: @@ -138,7 +138,7 @@ class Solution(object): else: hashmap[n1+n2] = 1 - # if the -(a+b) exists in nums3 and nums4, we shall add the count + # 如果 -(n1+n2) 存在于nums3和nums4, 存入结果 count = 0 for n3 in nums3: for n4 in nums4: @@ -149,20 +149,18 @@ class Solution(object): ``` - +（版本二）使用 defaultdict ```python +from collections import defaultdict class Solution: def fourSumCount(self, nums1: list, nums2: list, nums3: list, nums4: list) -> int: - from collections import defaultdict # You may use normal dict instead. rec, cnt = defaultdict(lambda : 0), 0 - # To store the summary of all the possible combinations of nums1 & nums2, together with their frequencies. for i in nums1: for j in nums2: rec[i+j] += 1 - # To add up the frequencies if the corresponding value occurs in the dictionary for i in nums3: for j in nums4: - cnt += rec.get(-(i+j), 0) # No matched key, return 0. + cnt += rec.get(-(i+j), 0) return cnt ``` From 5cff41ab09e2b3866f3245c3d06d61b3b432f485 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 20:53:12 -0500 Subject: [PATCH 101/154] =?UTF-8?q?Update=200454.=E5=9B=9B=E6=95=B0?= =?UTF-8?q?=E7=9B=B8=E5=8A=A0II.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0454.四数相加II.md | 24 +++++++++++++++++++++++- 1 file changed, 23 insertions(+), 1 deletion(-) diff --git a/problems/0454.四数相加II.md b/problems/0454.四数相加II.md index 7818c02b..31f0504f 100644 --- a/problems/0454.四数相加II.md +++ b/problems/0454.四数相加II.md @@ -149,7 +149,29 @@ class Solution(object): ``` -（版本二）使用 defaultdict +（版本二） 使用字典 +```python +class Solution(object): + def fourSumCount(self, nums1, nums2, nums3, nums4): + # 使用字典存储nums1和nums2中的元素及其和 + hashmap = dict() + for n1 in nums1: + for n2 in nums2: + hashmap[n1+n2] = hashmap.get(n1+n2, 0) + 1 + + # 如果 -(n1+n2) 存在于nums3和nums4, 存入结果 + count = 0 + for n3 in nums3: + for n4 in nums4: + key = - n3 - n4 + if key in hashmap: + count += hashmap[key] + return count + + + +``` +（版本三）使用 defaultdict ```python from collections import defaultdict class Solution: From 38bc7021c6f15136e32785398ad17fdab9098d2b Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:05:34 -0500 Subject: [PATCH 102/154] =?UTF-8?q?Update=200383.=E8=B5=8E=E9=87=91?= =?UTF-8?q?=E4=BF=A1.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0383.赎金信.md | 98 ++++++++++++++------------------------ 1 file changed, 37 insertions(+), 61 deletions(-) diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index d9a184b6..c2d80d8d 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -142,34 +142,27 @@ class Solution { ``` -Python写法一（使用数组作为哈希表）： - +（版本一）使用数组 ```python class Solution: def canConstruct(self, ransomNote: str, magazine: str) -> bool: - - arr = [0] * 26 - - for x in magazine: # 记录 magazine里各个字符出现次数 - arr[ord(x) - ord('a')] += 1 - - for x in ransomNote: # 在arr里对应的字符个数做--操作 - if arr[ord(x) - ord('a')] == 0: # 如果没有出现过直接返回 - return False - else: - arr[ord(x) - ord('a')] -= 1 - - return True + ransom_count = [0] * 26 + magazine_count = [0] * 26 + for c in ransomNote: + ransom_count[ord(c) - ord('a')] += 1 + for c in magazine: + magazine_count[ord(c) - ord('a')] += 1 + return all(ransom_count[i] <= magazine_count[i] for i in range(26)) ``` -Python写法二（使用defaultdict）： +（版本二）使用defaultdict ```python +from collections import defaultdict + class Solution: def canConstruct(self, ransomNote: str, magazine: str) -> bool: - from collections import defaultdict - hashmap = defaultdict(int) for x in magazine: @@ -177,59 +170,42 @@ class Solution: for x in ransomNote: value = hashmap.get(x) - if value is None or value == 0: + if not value or not value: return False else: hashmap[x] -= 1 return True ``` - -Python写法三： - -```python -class Solution(object): - def canConstruct(self, ransomNote, magazine): - """ - :type ransomNote: str - :type magazine: str - :rtype: bool - """ - - # use a dict to store the number of letter occurance in ransomNote - hashmap = dict() - for s in ransomNote: - if s in hashmap: - hashmap[s] += 1 - else: - hashmap[s] = 1 - - # check if the letter we need can be found in magazine - for l in magazine: - if l in hashmap: - hashmap[l] -= 1 - - for key in hashmap: - if hashmap[key] > 0: - return False - - return True -``` - -Python写法四： +（版本三）使用字典 ```python class Solution: def canConstruct(self, ransomNote: str, magazine: str) -> bool: - c1 = collections.Counter(ransomNote) - c2 = collections.Counter(magazine) - x = c1 - c2 - #x只保留值大于0的符号，当c1里面的符号个数小于c2时，不会被保留 - #所以x只保留下了，magazine不能表达的 - if(len(x)==0): - return True - else: - return False + counts = {} + for c in magazine: + counts[c] = counts.get(c, 0) + 1 + for c in ransomNote: + if c not in counts or counts[c] == 0: + return False + counts[c] -= 1 + return True +``` +（版本四）使用Counter + +```python +from collections import Counter + +class Solution: + def canConstruct(self, ransomNote: str, magazine: str) -> bool: + return not Counter(ransomNote) - Counter(magazine) +``` +（版本五）使用count + +```python +class Solution: + def canConstruct(self, ransomNote: str, magazine: str) -> bool: + return all(ransomNote.count(c) <= magazine.count(c) for c in set(ransomNote)) ``` Go： From 1bc667eccdb247b9795e28a8793ec51fe9ed4883 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:06:23 -0500 Subject: [PATCH 103/154] =?UTF-8?q?Update=200383.=E8=B5=8E=E9=87=91?= =?UTF-8?q?=E4=BF=A1.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0383.赎金信.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index c2d80d8d..75729437 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -206,6 +206,7 @@ class Solution: class Solution: def canConstruct(self, ransomNote: str, magazine: str) -> bool: return all(ransomNote.count(c) <= magazine.count(c) for c in set(ransomNote)) + ``` Go： From 2ddbe7f3404ca4c7e70cf3264b6ba4e02327ca78 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:30:24 -0500 Subject: [PATCH 104/154] =?UTF-8?q?Update=200015.=E4=B8=89=E6=95=B0?= =?UTF-8?q?=E4=B9=8B=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0015.三数之和.md | 95 +++++++++++++++++++---------------- 1 file changed, 53 insertions(+), 42 deletions(-) diff --git a/problems/0015.三数之和.md b/problems/0015.三数之和.md index 26c9eaa2..9cca779b 100644 --- a/problems/0015.三数之和.md +++ b/problems/0015.三数之和.md @@ -298,61 +298,72 @@ class Solution { ``` Python： +（版本一） 双指针 ```Python class Solution: - def threeSum(self, nums): - ans = [] - n = len(nums) + def threeSum(self, nums: List[int]) -> List[List[int]]: + result = [] nums.sort() - # 找出a + b + c = 0 - # a = nums[i], b = nums[left], c = nums[right] - for i in range(n): - left = i + 1 - right = n - 1 - # 排序之后如果第一个元素已经大于零，那么无论如何组合都不可能凑成三元组，直接返回结果就可以了 - if nums[i] > 0: - break - if i >= 1 and nums[i] == nums[i - 1]: # 去重a + + for i in range(len(nums)): + # 如果第一个元素已经大于0，不需要进一步检查 + if nums[i] > 0: + return result + + # 跳过相同的元素以避免重复 + if i > 0 and nums[i] == nums[i - 1]: continue - while left < right: - total = nums[i] + nums[left] + nums[right] - if total > 0: - right -= 1 - elif total < 0: + + left = i + 1 + right = len(nums) - 1 + + while right > left: + sum_ = nums[i] + nums[left] + nums[right] + + if sum_ < 0: left += 1 + elif sum_ > 0: + right -= 1 else: - ans.append([nums[i], nums[left], nums[right]]) - # 去重逻辑应该放在找到一个三元组之后，对b 和 c去重 - while left != right and nums[left] == nums[left + 1]: left += 1 - while left != right and nums[right] == nums[right - 1]: right -= 1 - left += 1 + result.append([nums[i], nums[left], nums[right]]) + + # 跳过相同的元素以避免重复 + while right > left and nums[right] == nums[right - 1]: + right -= 1 + while right > left and nums[left] == nums[left + 1]: + left += 1 + right -= 1 - return ans + left += 1 + + return result ``` -Python (v3): +（版本二） 使用字典 ```python class Solution: def threeSum(self, nums: List[int]) -> List[List[int]]: - if len(nums) < 3: return [] - nums, res = sorted(nums), [] - for i in range(len(nums) - 2): - cur, l, r = nums[i], i + 1, len(nums) - 1 - if res != [] and res[-1][0] == cur: continue # Drop duplicates for the first time. - - while l < r: - if cur + nums[l] + nums[r] == 0: - res.append([cur, nums[l], nums[r]]) - # Drop duplicates for the second time in interation of l & r. Only used when target situation occurs, because that is the reason for dropping duplicates. - while l < r - 1 and nums[l] == nums[l + 1]: - l += 1 - while r > l + 1 and nums[r] == nums[r - 1]: - r -= 1 - if cur + nums[l] + nums[r] > 0: - r -= 1 + result = [] + nums.sort() + # 找出a + b + c = 0 + # a = nums[i], b = nums[j], c = -(a + b) + for i in range(len(nums)): + # 排序之后如果第一个元素已经大于零，那么不可能凑成三元组 + if nums[i] > 0: + break + if i > 0 and nums[i] == nums[i - 1]: #三元组元素a去重 + continue + d = {} + for j in range(i + 1, len(nums)): + if j > i + 2 and nums[j] == nums[j-1] == nums[j-2]: # 三元组元素b去重 + continue + c = 0 - (nums[i] + nums[j]) + if c in d: + result.append([nums[i], nums[j], c]) + d.pop(c) # 三元组元素c去重 else: - l += 1 - return res + d[nums[j]] = j + return result ``` Go： From a5bb942c95f405181ba311e14e1ad601bcfbf435 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:35:49 -0500 Subject: [PATCH 105/154] =?UTF-8?q?Update=200018.=E5=9B=9B=E6=95=B0?= =?UTF-8?q?=E4=B9=8B=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0018.四数之和.md | 53 +++++++++++++++++++---------------- 1 file changed, 29 insertions(+), 24 deletions(-) diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 5f4c2ec9..8606b2eb 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -207,35 +207,40 @@ class Solution { ``` Python： +(版本一) 双指针 ```python -# 双指针法 -class Solution: - def fourSum(self, nums: List[int], target: int) -> List[List[int]]: - + def fourSum(self, nums: List[int], target: int) -> List[List[int]]: + result = [] nums.sort() - n = len(nums) - res = [] - for i in range(n): - if i > 0 and nums[i] == nums[i - 1]: continue # 对nums[i]去重 - for k in range(i+1, n): - if k > i + 1 and nums[k] == nums[k-1]: continue # 对nums[k]去重 - p = k + 1 - q = n - 1 - - while p < q: - if nums[i] + nums[k] + nums[p] + nums[q] > target: q -= 1 - elif nums[i] + nums[k] + nums[p] + nums[q] < target: p += 1 + for k in range(len(nums)): + if nums[k] > target and nums[k] >= 0: + break + if k > 0 and nums[k] == nums[k-1]: + continue + for i in range(k+1, len(nums)): + if nums[k] + nums[i] > target and nums[k] + nums[i] >= 0: + break + if i > k+1 and nums[i] == nums[i-1]: + continue + left, right = i+1, len(nums)-1 + while right > left: + if nums[k] + nums[i] + nums[left] + nums[right] > target: + right -= 1 + elif nums[k] + nums[i] + nums[left] + nums[right] < target: + left += 1 else: - res.append([nums[i], nums[k], nums[p], nums[q]]) - # 对nums[p]和nums[q]去重 - while p < q and nums[p] == nums[p + 1]: p += 1 - while p < q and nums[q] == nums[q - 1]: q -= 1 - p += 1 - q -= 1 - return res + result.append([nums[k], nums[i], nums[left], nums[right]]) + while right > left and nums[right] == nums[right-1]: + right -= 1 + while right > left and nums[left] == nums[left+1]: + left += 1 + right -= 1 + left += 1 + return result ``` +(版本二) 使用字典 + ```python -# 哈希表法 class Solution(object): def fourSum(self, nums, target): """ From dde92b1da3808200925d82f8f523da7b5cd93ca2 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:36:38 -0500 Subject: [PATCH 106/154] =?UTF-8?q?Update=200018.=E5=9B=9B=E6=95=B0?= =?UTF-8?q?=E4=B9=8B=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0018.四数之和.md | 40 +++++++++++++++++++---------------- 1 file changed, 22 insertions(+), 18 deletions(-) diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 8606b2eb..4fee8358 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -209,34 +209,38 @@ class Solution { Python： (版本一) 双指针 ```python - def fourSum(self, nums: List[int], target: int) -> List[List[int]]: - result = [] +class Solution: + def fourSum(self, nums: List[int], target: int) -> List[List[int]]: nums.sort() - for k in range(len(nums)): - if nums[k] > target and nums[k] >= 0: + n = len(nums) + result = [] + for i in range(n): + if nums[i] > target and nums[i] > 0 and target > 0:# 剪枝（可省） break - if k > 0 and nums[k] == nums[k-1]: + if i > 0 and nums[i] == nums[i-1]:# 去重 continue - for i in range(k+1, len(nums)): - if nums[k] + nums[i] > target and nums[k] + nums[i] >= 0: + for j in range(i+1, n): + if nums[i] + nums[j] > target and target > 0: #剪枝（可省） break - if i > k+1 and nums[i] == nums[i-1]: + if j > i+1 and nums[j] == nums[j-1]: # 去重 continue - left, right = i+1, len(nums)-1 - while right > left: - if nums[k] + nums[i] + nums[left] + nums[right] > target: + left, right = j+1, n-1 + while left < right: + s = nums[i] + nums[j] + nums[left] + nums[right] + if s == target: + result.append([nums[i], nums[j], nums[left], nums[right]]) + while left < right and nums[left] == nums[left+1]: + left += 1 + while left < right and nums[right] == nums[right-1]: + right -= 1 + left += 1 right -= 1 - elif nums[k] + nums[i] + nums[left] + nums[right] < target: + elif s < target: left += 1 else: - result.append([nums[k], nums[i], nums[left], nums[right]]) - while right > left and nums[right] == nums[right-1]: - right -= 1 - while right > left and nums[left] == nums[left+1]: - left += 1 right -= 1 - left += 1 return result + ``` (版本二) 使用字典 From c4bc3197502473c5f14be392157bb83ddaa2dbeb Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Fri, 5 May 2023 21:39:52 -0500 Subject: [PATCH 107/154] =?UTF-8?q?Update=200018.=E5=9B=9B=E6=95=B0?= =?UTF-8?q?=E4=B9=8B=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0018.四数之和.md | 33 +++++++++++---------------------- 1 file changed, 11 insertions(+), 22 deletions(-) diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 4fee8358..a4d41d9b 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -252,36 +252,25 @@ class Solution(object): :type target: int :rtype: List[List[int]] """ - # use a dict to store value:showtimes - hashmap = dict() - for n in nums: - if n in hashmap: - hashmap[n] += 1 - else: - hashmap[n] = 1 + # 创建一个字典来存储输入列表中每个数字的频率 + freq = {} + for num in nums: + freq[num] = freq.get(num, 0) + 1 - # good thing about using python is you can use set to drop duplicates. + # 创建一个集合来存储最终答案，并遍历4个数字的所有唯一组合 ans = set() - # ans = [] # save results by list() for i in range(len(nums)): for j in range(i + 1, len(nums)): for k in range(j + 1, len(nums)): val = target - (nums[i] + nums[j] + nums[k]) - if val in hashmap: - # make sure no duplicates. + if val in freq: + # 确保没有重复 count = (nums[i] == val) + (nums[j] == val) + (nums[k] == val) - if hashmap[val] > count: - ans_tmp = tuple(sorted([nums[i], nums[j], nums[k], val])) - ans.add(ans_tmp) - # Avoiding duplication in list manner but it cause time complexity increases - # if ans_tmp not in ans: - # ans.append(ans_tmp) - else: - continue - return list(ans) - # if used list() to save results, just - # return ans + if freq[val] > count: + ans.add(tuple(sorted([nums[i], nums[j], nums[k], val]))) + return [list(x) for x in ans] + ``` Go： From b617532ce9b7af1cd163a0ccccb1f956f77e8cd2 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 01:55:48 -0500 Subject: [PATCH 108/154] =?UTF-8?q?Update=200513.=E6=89=BE=E6=A0=91?= =?UTF-8?q?=E5=B7=A6=E4=B8=8B=E8=A7=92=E7=9A=84=E5=80=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0513.找树左下角的值.md | 102 ++++++++++++++++--------- 1 file changed, 65 insertions(+), 37 deletions(-) diff --git a/problems/0513.找树左下角的值.md b/problems/0513.找树左下角的值.md index 5ff25484..26768c74 100644 --- a/problems/0513.找树左下角的值.md +++ b/problems/0513.找树左下角的值.md @@ -271,52 +271,80 @@ class Solution { ### Python - -递归: +（版本一）递归法 + 回溯 ```python class Solution: def findBottomLeftValue(self, root: TreeNode) -> int: - max_depth = -float("INF") - leftmost_val = 0 + self.max_depth = float('-inf') + self.result = None + self.traversal(root, 0) + return self.result + + def traversal(self, node, depth): + if not node.left and not node.right: + if depth > self.max_depth: + self.max_depth = depth + self.result = node.val + return + + if node.left: + depth += 1 + self.traversal(node.left, depth) + depth -= 1 + if node.right: + depth += 1 + self.traversal(node.right, depth) + depth -= 1 - def __traverse(root, cur_depth): - nonlocal max_depth, leftmost_val - if not root.left and not root.right: - if cur_depth > max_depth: - max_depth = cur_depth - leftmost_val = root.val - if root.left: - cur_depth += 1 - __traverse(root.left, cur_depth) - cur_depth -= 1 - if root.right: - cur_depth += 1 - __traverse(root.right, cur_depth) - cur_depth -= 1 - - __traverse(root, 0) - return leftmost_val ``` -迭代 - 层序遍历: +（版本二）递归法+精简 ```python class Solution: def findBottomLeftValue(self, root: TreeNode) -> int: - queue = deque() - if root: - queue.append(root) - result = 0 - while queue: - q_len = len(queue) - for i in range(q_len): - if i == 0: - result = queue[i].val - cur = queue.popleft() - if cur.left: - queue.append(cur.left) - if cur.right: - queue.append(cur.right) - return result + self.max_depth = float('-inf') + self.result = None + self.traversal(root, 0) + return self.result + + def traversal(self, node, depth): + if not node.left and not node.right: + if depth > self.max_depth: + self.max_depth = depth + self.result = node.val + return + + if node.left: + self.traversal(node.left, depth+1) + if node.right: + self.traversal(node.right, depth+1) +``` + +(版本三） 迭代法 +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +from collections import deque +class Solution: + def findBottomLeftValue(self, root: Optional[TreeNode]) -> int: + queue = deque([root]) + while queue: + size = len(queue) + leftmost = queue[0].val + for i in range(size): + node = queue.popleft() + if node.left: + queue.append(node.left) + if node.right: + queue.append(node.right) + if not queue: + return leftmost + + ``` ### Go From bb8b112ae0079dd0d09d6a075e17888616ba3a55 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 16:51:15 -0500 Subject: [PATCH 109/154] =?UTF-8?q?Update=200344.=E5=8F=8D=E8=BD=AC?= =?UTF-8?q?=E5=AD=97=E7=AC=A6=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0344.反转字符串.md | 58 +++++++++++++++++++++++++++++++- 1 file changed, 57 insertions(+), 1 deletion(-) diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md index 775cfc58..03c82767 100644 --- a/problems/0344.反转字符串.md +++ b/problems/0344.反转字符串.md @@ -174,6 +174,7 @@ class Solution { ``` Python： +（版本一） 双指针 ```python class Solution: def reverseString(self, s: List[str]) -> None: @@ -190,7 +191,62 @@ class Solution: right -= 1 ``` - +（版本二） 使用栈 +```python +class Solution: + def reverseString(self, s: List[str]) -> None: + """ + Do not return anything, modify s in-place instead. + """ + stack = [] + for char in s: + stack.append(char) + for i in range(len(s)): + s[i] = stack.pop() + +``` +（版本三） 使用range +```python +class Solution: + def reverseString(self, s: List[str]) -> None: + """ + Do not return anything, modify s in-place instead. + """ + n = len(s) + for i in range(n // 2): + s[i], s[n - i - 1] = s[n - i - 1], s[i] + +``` +（版本四） 使用reversed +```python +class Solution: + def reverseString(self, s: List[str]) -> None: + """ + Do not return anything, modify s in-place instead. + """ + s[:] = reversed(s) + +``` +（版本五） 使用切片 +```python +class Solution: + def reverseString(self, s: List[str]) -> None: + """ + Do not return anything, modify s in-place instead. + """ + s[:] = s[::-1] + +``` +（版本六） 使用列表推导 +```python +class Solution: + def reverseString(self, s: List[str]) -> None: + """ + Do not return anything, modify s in-place instead. + """ + s[:] = [s[i] for i in range(len(s) - 1, -1, -1)] + +``` Go： ```Go func reverseString(s []byte) { From a37ae94618edd6104f9d0815bc0d97e185175f1e Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 16:54:34 -0500 Subject: [PATCH 110/154] =?UTF-8?q?Update=200344.=E5=8F=8D=E8=BD=AC?= =?UTF-8?q?=E5=AD=97=E7=AC=A6=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0344.反转字符串.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/0344.反转字符串.md b/problems/0344.反转字符串.md index 03c82767..1c74f9aa 100644 --- a/problems/0344.反转字符串.md +++ b/problems/0344.反转字符串.md @@ -183,8 +183,8 @@ class Solution: """ left, right = 0, len(s) - 1 - # 该方法已经不需要判断奇偶数，经测试后时间空间复杂度比用 for i in range(right//2)更低 - # 推荐该写法，更加通俗易懂 + # 该方法已经不需要判断奇偶数，经测试后时间空间复杂度比用 for i in range(len(s)//2)更低 + # 因为while每次循环需要进行条件判断，而range函数不需要，直接生成数字，因此时间复杂度更低。推荐使用range while left < right: s[left], s[right] = s[right], s[left] left += 1 From abcedd41a87e6923cfd39705765cdd5716b31ad0 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 16:59:12 -0500 Subject: [PATCH 111/154] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer05.?= =?UTF-8?q?=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index 2e1ee1de..833d4628 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -266,6 +266,7 @@ func replaceSpace(s string) string { python： +####因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) ```python class Solution: def replaceSpace(self, s: str) -> str: From 3958cc166a1ab090f0060e45bab41535170cd186 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 16:59:48 -0500 Subject: [PATCH 112/154] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer05.?= =?UTF-8?q?=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index 833d4628..d930702f 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -265,8 +265,8 @@ func replaceSpace(s string) string { -python： -####因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) +#python： +#因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) ```python class Solution: def replaceSpace(self, s: str) -> str: From 799325bacebd60e3088e5ab4a82ef599ba1d3e47 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 17:00:54 -0500 Subject: [PATCH 113/154] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer05.?= =?UTF-8?q?=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index d930702f..a6c32d30 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -265,8 +265,8 @@ func replaceSpace(s string) string { -#python： -#因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) +# python： +# 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) ```python class Solution: def replaceSpace(self, s: str) -> str: From dc7ac6337d5ae0a462cc332a5a1a68de43095212 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 17:01:39 -0500 Subject: [PATCH 114/154] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer05.?= =?UTF-8?q?=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index a6c32d30..d86cc6dc 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -265,8 +265,8 @@ func replaceSpace(s string) string { -# python： -# 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) +python： +#### 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) ```python class Solution: def replaceSpace(self, s: str) -> str: From c09089accda7c3cc46bc283eb63edfc9563894ab Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 17:10:18 -0500 Subject: [PATCH 115/154] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer05.?= =?UTF-8?q?=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 34 ++++++++++++++++++++------ 1 file changed, 27 insertions(+), 7 deletions(-) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index d86cc6dc..dbad781e 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -266,7 +266,8 @@ func replaceSpace(s string) string { python： -#### 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能不为O(1) +#### 因为字符串是不可变类型，所以操作字符串需要将其转换为列表，因此空间复杂度不可能为O(1) +（版本一）转换成列表，并且添加相匹配的空间，然后进行填充 ```python class Solution: def replaceSpace(self, s: str) -> str: @@ -291,14 +292,22 @@ class Solution: return ''.join(res) ``` - +（版本二）添加空列表，添加匹配的结果 +```python +class Solution: + def replaceSpace(self, s: str) -> str: + res = [] + for i in range(len(s)): + if s[i] == ' ': + res.append('%20') + else: + res.append(s[i]) + return ''.join(res) +``` +（版本三）使用切片 ```python class Solution: def replaceSpace(self, s: str) -> str: - # method 1 - Very rude - return "%20".join(s.split(" ")) - - # method 2 - Reverse the s when counting in for loop, then update from the end. n = len(s) for e, i in enumerate(s[::-1]): print(i, e) @@ -307,7 +316,18 @@ class Solution: print("") return s ``` - +（版本四）使用join + split +```python +class Solution: + def replaceSpace(self, s: str) -> str: + return "%20".join(s.split(" ")) +``` +（版本五）使用replace +```python +class Solution: + def replaceSpace(self, s: str) -> str: + return s.replace(' ', '%20') +``` javaScript: ```js From c75140d2572be2030e64a8aa18a7cc336b7a5a74 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 17:26:08 -0500 Subject: [PATCH 116/154] =?UTF-8?q?Update=200151.=E7=BF=BB=E8=BD=AC?= =?UTF-8?q?=E5=AD=97=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95=E8=AF=8D?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0151.翻转字符串里的单词.md | 136 +++---------------- 1 file changed, 21 insertions(+), 115 deletions(-) diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md index 8fa7c77c..a3232264 100644 --- a/problems/0151.翻转字符串里的单词.md +++ b/problems/0151.翻转字符串里的单词.md @@ -434,134 +434,40 @@ class Solution { ``` python: - +（版本一）先删除空白，然后整个反转，最后单词反转。 +### 因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1) ```Python class Solution: - #1.去除多余的空格 - def trim_spaces(self, s): - n = len(s) - left = 0 - right = n-1 - - while left <= right and s[left] == ' ': #去除开头的空格 - left += 1 - while left <= right and s[right] == ' ': #去除结尾的空格 - right -= 1 - tmp = [] - while left <= right: #去除单词中间多余的空格 - if s[left] != ' ': - tmp.append(s[left]) - elif tmp[-1] != ' ': #当前位置是空格，但是相邻的上一个位置不是空格，则该空格是合理的 - tmp.append(s[left]) - left += 1 - return tmp - - #2.翻转字符数组 - def reverse_string(self, nums, left, right): - while left < right: - nums[left], nums[right] = nums[right], nums[left] - left += 1 - right -= 1 - return None - - #3.翻转每个单词 - def reverse_each_word(self, nums): - start = 0 - end = 0 - n = len(nums) - while start < n: - while end < n and nums[end] != ' ': - end += 1 - self.reverse_string(nums, start, end-1) - start = end + 1 - end += 1 - return None - -#4.翻转字符串里的单词 - def reverseWords(self, s): #测试用例："the sky is blue" - l = self.trim_spaces(s) #输出：['t', 'h', 'e', ' ', 's', 'k', 'y', ' ', 'i', 's', ' ', 'b', 'l', 'u', 'e' - self.reverse_string(l, 0, len(l)-1) #输出：['e', 'u', 'l', 'b', ' ', 's', 'i', ' ', 'y', 'k', 's', ' ', 'e', 'h', 't'] - self.reverse_each_word(l) #输出：['b', 'l', 'u', 'e', ' ', 'i', 's', ' ', 's', 'k', 'y', ' ', 't', 'h', 'e'] - return ''.join(l) #输出：blue is sky the - + def reverseWords(self, s: str) -> str: + # 删除前后空白 + s = s.strip() + # 反转整个字符串 + s = s[::-1] + # 将字符串拆分为单词，并反转每个单词 + s = ' '.join(word[::-1] for word in s.split()) + return s ``` +（版本二）使用双指针 ```python class Solution: def reverseWords(self, s: str) -> str: - # method 1 - Rude but work & efficient method. - s_list = [i for i in s.split(" ") if len(i) > 0] - return " ".join(s_list[::-1]) + # 将字符串拆分为单词，即转换成列表类型 + words = s.split() - # method 2 - Carlo's idea - def trim_head_tail_space(ss: str): - p = 0 - while p < len(ss) and ss[p] == " ": - p += 1 - return ss[p:] + # 反转单词 + left, right = 0, len(words) - 1 + while left < right: + words[left], words[right] = words[right], words[left] + left += 1 + right -= 1 - # Trim the head and tail space - s = trim_head_tail_space(s) - s = trim_head_tail_space(s[::-1])[::-1] - - pf, ps, s = 0, 0, s[::-1] # Reverse the string. - while pf < len(s): - if s[pf] == " ": - # Will not excede. Because we have clean the tail space. - if s[pf] == s[pf + 1]: - s = s[:pf] + s[pf + 1:] - continue - else: - s = s[:ps] + s[ps: pf][::-1] + s[pf:] - ps, pf = pf + 1, pf + 2 - else: - pf += 1 - return s[:ps] + s[ps:][::-1] # Must do the last step, because the last word is omit though the pointers are on the correct positions, + # 将列表转换成字符串 + return " ".join(words) ``` -```python -class Solution: # 使用双指针法移除空格 - def reverseWords(self, s: str) -> str: - - def removeextraspace(s): - start = 0; end = len(s)-1 - while s[start]==' ': - start+=1 - while s[end]==' ': - end-=1 - news = list(s[start:end+1]) - slow = fast = 0 - while fast0 and news[fast]==news[fast-1]==' ': - fast+=1 - news[slow]=news[fast] - slow+=1; fast+=1 - #return "".join(news[:slow]) - return news[:slow] - def reversestr(s): - left,right = 0,len(s)-1 - news = list(s) - while left Date: Sat, 6 May 2023 17:26:56 -0500 Subject: [PATCH 117/154] =?UTF-8?q?Update=200151.=E7=BF=BB=E8=BD=AC?= =?UTF-8?q?=E5=AD=97=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95=E8=AF=8D?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0151.翻转字符串里的单词.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md index a3232264..062ccca5 100644 --- a/problems/0151.翻转字符串里的单词.md +++ b/problems/0151.翻转字符串里的单词.md @@ -435,7 +435,7 @@ class Solution { python: （版本一）先删除空白，然后整个反转，最后单词反转。 -### 因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1) +#### 因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1) ```Python class Solution: def reverseWords(self, s: str) -> str: From 7cfd38940249adb1f659da29b69edc2ae3083113 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 17:52:08 -0500 Subject: [PATCH 118/154] =?UTF-8?q?Update=20=E5=89=91=E6=8C=87Offer58-II.?= =?UTF-8?q?=E5=B7=A6=E6=97=8B=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../剑指Offer58-II.左旋转字符串.md | 54 ++++++++++--------- 1 file changed, 28 insertions(+), 26 deletions(-) diff --git a/problems/剑指Offer58-II.左旋转字符串.md b/problems/剑指Offer58-II.左旋转字符串.md index f368514f..6cd88456 100644 --- a/problems/剑指Offer58-II.左旋转字符串.md +++ b/problems/剑指Offer58-II.左旋转字符串.md @@ -142,15 +142,16 @@ class Solution { ``` python: +（版本一）使用切片 ```python -# 方法一：可以使用切片方法 class Solution: def reverseLeftWords(self, s: str, n: int) -> str: - return s[n:] + s[0:n] + return s[n:] + s[:n] ``` +（版本二）使用reversed + join + ```python -# 方法二：也可以使用上文描述的方法，有些面试中不允许使用切片，那就使用上文作者提到的方法 class Solution: def reverseLeftWords(self, s: str, n: int) -> str: s = list(s) @@ -161,32 +162,29 @@ class Solution: return "".join(s) ``` +（版本三）自定义reversed函数 ```python -# 方法三：如果连reversed也不让使用，那么自己手写一个 class Solution: - def reverseLeftWords(self, s: str, n: int) -> str: - def reverse_sub(lst, left, right): - while left < right: - lst[left], lst[right] = lst[right], lst[left] - left += 1 - right -= 1 + def reverseLeftWords(self, s: str, n: int) -> str: + s_list = list(s) - res = list(s) - end = len(res) - 1 - reverse_sub(res, 0, n - 1) - reverse_sub(res, n, end) - reverse_sub(res, 0, end) - return ''.join(res) + self.reverse(s_list, 0, n - 1) + self.reverse(s_list, n, len(s_list) - 1) + self.reverse(s_list, 0, len(s_list) - 1) -# 同方法二 -# 时间复杂度：O(n) -# 空间复杂度：O(n)，python的string为不可变，需要开辟同样大小的list空间来修改 + return ''.join(s_list) + + def reverse(self, s, start, end): + while start < end: + s[start], s[end] = s[end], s[start] + start += 1 + end -= 1 ``` +（版本四）使用 模 +下标 ```python 3 -#方法四：考虑不能用切片的情况下，利用模+下标实现 class Solution: def reverseLeftWords(self, s: str, n: int) -> str: new_s = '' @@ -196,17 +194,21 @@ class Solution: return new_s ``` +（版本五）使用 模 + 切片 ```python 3 -# 方法五：另类的切片方法 class Solution: def reverseLeftWords(self, s: str, n: int) -> str: - n = len(s) - s = s + s - return s[k : n+k] + l = len(s) + # 复制输入字符串与它自己连接 + s = s + s + + # 计算旋转字符串的起始索引 + k = n % (l * 2) + + # 从连接的字符串中提取旋转后的字符串并返回 + return s[k : k + l] -# 时间复杂度：O(n) -# 空间复杂度：O(n) ``` Go： From 3cf55028680832e76c82f9f819443b79339f1d49 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 19:39:34 -0500 Subject: [PATCH 119/154] =?UTF-8?q?Update=200028.=E5=AE=9E=E7=8E=B0strStr.?= =?UTF-8?q?md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0028.实现strStr.md | 99 +++++++++++++++++++++-------------- 1 file changed, 59 insertions(+), 40 deletions(-) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index 263c1689..5cc04994 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -692,23 +692,38 @@ class Solution { ``` Python3： +（版本一）前缀表（不减一） + ```python -//暴力解法： -class Solution(object): - def strStr(self, haystack, needle): - """ - :type haystack: str - :type needle: str - :rtype: int - """ - m, n = len(haystack), len(needle) - for i in range(m): - if haystack[i:i+n] == needle: - return i - return -1 +class Solution: + def strStr(self, haystack: str, needle: str) -> int: + if len(needle) == 0: + return 0 + next = self.getNext(needle) + j = 0 + for i in range(len(haystack)): + while j >= 1 and haystack[i] != needle[j]: + j = next[j-1] + if haystack[i] == needle[j]: + j += 1 + if j == len(needle): + return i - len(needle) + 1 + return -1 + + def getNext(self, needle): + next = [0] * len(needle) + j = 0 + next[0] = j + for i in range(1, len(needle)): + while j >= 1 and needle[i] != needle[j]: + j = next[j-1] + if needle[i] == needle[j]: + j += 1 + next[i] = j + return next ``` +（版本二）前缀表（减一） ```python -// 方法一 class Solution: def strStr(self, haystack: str, needle: str) -> int: a = len(needle) @@ -739,8 +754,11 @@ class Solution: return next ``` + +（版本三）前缀表（减一） + + ```python -// 方法二 class Solution: def strStr(self, haystack: str, needle: str) -> int: a = len(needle) @@ -774,35 +792,36 @@ class Solution: return next ``` +（版本四）暴力法 +```python +class Solution(object): + def strStr(self, haystack, needle): + """ + :type haystack: str + :type needle: str + :rtype: int + """ + m, n = len(haystack), len(needle) + for i in range(m): + if haystack[i:i+n] == needle: + return i + return -1 +``` +（版本五）使用 index ```python -// 前缀表（不减一）Python实现 class Solution: def strStr(self, haystack: str, needle: str) -> int: - if len(needle) == 0: - return 0 - next = self.getNext(needle) - j = 0 - for i in range(len(haystack)): - while j >= 1 and haystack[i] != needle[j]: - j = next[j-1] - if haystack[i] == needle[j]: - j += 1 - if j == len(needle): - return i - len(needle) + 1 - return -1 - - def getNext(self, needle): - next = [0] * len(needle) - j = 0 - next[0] = j - for i in range(1, len(needle)): - while j >= 1 and needle[i] != needle[j]: - j = next[j-1] - if needle[i] == needle[j]: - j += 1 - next[i] = j - return next + try: + return haystack.index(needle) + except ValueError: + return -1 ``` +（版本六）使用 find +```python +class Solution: + def strStr(self, haystack: str, needle: str) -> int: + return haystack.find(needle) + Go： From 1fbd900c6a942b86d9c1d2f394ed89023e22a50d Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 19:47:55 -0500 Subject: [PATCH 120/154] =?UTF-8?q?Update=200028.=E5=AE=9E=E7=8E=B0strStr.?= =?UTF-8?q?md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0028.实现strStr.md | 133 ++++++++++++---------------------- 1 file changed, 46 insertions(+), 87 deletions(-) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index 5cc04994..ff13db39 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -692,107 +692,66 @@ class Solution { ``` Python3： -（版本一）前缀表（不减一） +（版本一）前缀表（减一） +```python +class Solution: + def getNext(self, next, s): + j = -1 + next[0] = j + for i in range(1, len(s)): + while j >= 0 and s[i] != s[j+1]: + j = next[j] + if s[i] == s[j+1]: + j += 1 + next[i] = j + + def strStr(self, haystack: str, needle: str) -> int: + if not needle: + return 0 + next = [0] * len(needle) + self.getNext(next, needle) + j = -1 + for i in range(len(haystack)): + while j >= 0 and haystack[i] != needle[j+1]: + j = next[j] + if haystack[i] == needle[j+1]: + j += 1 + if j == len(needle) - 1: + return i - len(needle) + 1 + return -1 +``` +（版本二）前缀表（不减一） ```python class Solution: + def getNext(self, next: List[int], s: str) -> None: + j = 0 + next[0] = 0 + for i in range(1, len(s)): + while j > 0 and s[i] != s[j]: + j = next[j - 1] + if s[i] == s[j]: + j += 1 + next[i] = j + def strStr(self, haystack: str, needle: str) -> int: if len(needle) == 0: return 0 - next = self.getNext(needle) + next = [0] * len(needle) + self.getNext(next, needle) j = 0 for i in range(len(haystack)): - while j >= 1 and haystack[i] != needle[j]: - j = next[j-1] + while j > 0 and haystack[i] != needle[j]: + j = next[j - 1] if haystack[i] == needle[j]: j += 1 if j == len(needle): return i - len(needle) + 1 return -1 - - def getNext(self, needle): - next = [0] * len(needle) - j = 0 - next[0] = j - for i in range(1, len(needle)): - while j >= 1 and needle[i] != needle[j]: - j = next[j-1] - if needle[i] == needle[j]: - j += 1 - next[i] = j - return next -``` -（版本二）前缀表（减一） -```python -class Solution: - def strStr(self, haystack: str, needle: str) -> int: - a = len(needle) - b = len(haystack) - if a == 0: - return 0 - next = self.getnext(a,needle) - p=-1 - for j in range(b): - while p >= 0 and needle[p+1] != haystack[j]: - p = next[p] - if needle[p+1] == haystack[j]: - p += 1 - if p == a-1: - return j-a+1 - return -1 - - def getnext(self,a,needle): - next = ['' for i in range(a)] - k = -1 - next[0] = k - for i in range(1, len(needle)): - while (k > -1 and needle[k+1] != needle[i]): - k = next[k] - if needle[k+1] == needle[i]: - k += 1 - next[i] = k - return next ``` -（版本三）前缀表（减一） - - -```python -class Solution: - def strStr(self, haystack: str, needle: str) -> int: - a = len(needle) - b = len(haystack) - if a == 0: - return 0 - i = j = 0 - next = self.getnext(a, needle) - while(i < b and j < a): - if j == -1 or needle[j] == haystack[i]: - i += 1 - j += 1 - else: - j = next[j] - if j == a: - return i-j - else: - return -1 - - def getnext(self, a, needle): - next = ['' for i in range(a)] - j, k = 0, -1 - next[0] = k - while(j < a-1): - if k == -1 or needle[k] == needle[j]: - k += 1 - j += 1 - next[j] = k - else: - k = next[k] - return next -``` - -（版本四）暴力法 +（版本三）暴力法 ```python class Solution(object): def strStr(self, haystack, needle): @@ -807,7 +766,7 @@ class Solution(object): return i return -1 ``` -（版本五）使用 index +（版本四）使用 index ```python class Solution: def strStr(self, haystack: str, needle: str) -> int: @@ -816,7 +775,7 @@ class Solution: except ValueError: return -1 ``` -（版本六）使用 find +（版本五）使用 find ```python class Solution: def strStr(self, haystack: str, needle: str) -> int: From 995028d7c45099f71d9ef398ad92a8b655e0c4f9 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 19:48:45 -0500 Subject: [PATCH 121/154] =?UTF-8?q?Update=200028.=E5=AE=9E=E7=8E=B0strStr.?= =?UTF-8?q?md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0028.实现strStr.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index ff13db39..b18131f3 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -780,7 +780,7 @@ class Solution: class Solution: def strStr(self, haystack: str, needle: str) -> int: return haystack.find(needle) - +``` Go： From e8b4db9d7a5cf58e56e58254195cc2a98d926479 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 19:49:19 -0500 Subject: [PATCH 122/154] =?UTF-8?q?Update=200028.=E5=AE=9E=E7=8E=B0strStr.?= =?UTF-8?q?md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0028.实现strStr.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/0028.实现strStr.md b/problems/0028.实现strStr.md index b18131f3..86308b47 100644 --- a/problems/0028.实现strStr.md +++ b/problems/0028.实现strStr.md @@ -780,6 +780,7 @@ class Solution: class Solution: def strStr(self, haystack: str, needle: str) -> int: return haystack.find(needle) + ``` Go： From 5000a978fd41dd32a639935459365fd58e691c37 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 20:05:02 -0500 Subject: [PATCH 123/154] =?UTF-8?q?Update=200459.=E9=87=8D=E5=A4=8D?= =?UTF-8?q?=E7=9A=84=E5=AD=90=E5=AD=97=E7=AC=A6=E4=B8=B2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0459.重复的子字符串.md | 39 ++++++++++++++++++++++++-- 1 file changed, 36 insertions(+), 3 deletions(-) diff --git a/problems/0459.重复的子字符串.md b/problems/0459.重复的子字符串.md index 98c02a25..5d56ad18 100644 --- a/problems/0459.重复的子字符串.md +++ b/problems/0459.重复的子字符串.md @@ -264,8 +264,7 @@ class Solution { Python： -这里使用了前缀表统一减一的实现方式 - +（版本一） 前缀表 减一 ```python class Solution: def repeatedSubstringPattern(self, s: str) -> bool: @@ -289,7 +288,7 @@ class Solution: return nxt ``` -前缀表（不减一）的代码实现 +（版本二） 前缀表 不减一 ```python class Solution: @@ -314,6 +313,40 @@ class Solution: return nxt ``` + +（版本三） 使用 find + +```python +class Solution: + def repeatedSubstringPattern(self, s: str) -> bool: + n = len(s) + if n <= 1: + return False + ss = s[1:] + s[:-1] + print(ss.find(s)) + return ss.find(s) != -1 +``` + +（版本四） 暴力法 + +```python +class Solution: + def repeatedSubstringPattern(self, s: str) -> bool: + n = len(s) + if n <= 1: + return False + + substr = "" + for i in range(1, n//2 + 1): + if n % i == 0: + substr = s[:i] + if substr * (n//i) == s: + return True + + return False +``` + + Go： 这里使用了前缀表统一减一的实现方式 From f97ff3a1beeb938107bd76dfa38dac3c93430095 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= Date: Sun, 7 May 2023 09:28:28 +0800 Subject: [PATCH 124/154] =?UTF-8?q?Update=200151.=E7=BF=BB=E8=BD=AC?= =?UTF-8?q?=E5=AD=97=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95=E8=AF=8D?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0151.翻转字符串里的单词.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0151.翻转字符串里的单词.md b/problems/0151.翻转字符串里的单词.md index 062ccca5..4474f1c6 100644 --- a/problems/0151.翻转字符串里的单词.md +++ b/problems/0151.翻转字符串里的单词.md @@ -435,7 +435,7 @@ class Solution { python: （版本一）先删除空白，然后整个反转，最后单词反转。 -#### 因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1) +**因为字符串是不可变类型，所以反转单词的时候，需要将其转换成列表，然后通过join函数再将其转换成列表，所以空间复杂度不是O(1)** ```Python class Solution: def reverseWords(self, s: str) -> str: From 3f45b56cef522ca54b648ded97a8d3b0c6a0f919 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= Date: Sun, 7 May 2023 09:47:01 +0800 Subject: [PATCH 125/154] =?UTF-8?q?Update=200242.=E6=9C=89=E6=95=88?= =?UTF-8?q?=E7=9A=84=E5=AD=97=E6=AF=8D=E5=BC=82=E4=BD=8D=E8=AF=8D.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0242.有效的字母异位词.md | 68 +---------------------- 1 file changed, 1 insertion(+), 67 deletions(-) diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md index 0e5683c7..3f4b00dc 100644 --- a/problems/0242.有效的字母异位词.md +++ b/problems/0242.有效的字母异位词.md @@ -122,21 +122,7 @@ class Solution { ``` Python： -(版本一) 使用数组 -```python -class Solution: - def isAnagram(self, s: str, t: str) -> bool: - record = [0] * 26 - for i in range(len(s)): - record[ord(s[i]) - ord("a")] += 1 - for i in range(len(t)): - record[ord(t[i]) - ord("a")] -= 1 - for i in range(26): - if record[i] != 0: - return False - return True -``` -(版本二) 使用数组 + ```python class Solution: def isAnagram(self, s: str, t: str) -> bool: @@ -153,59 +139,7 @@ class Solution: return True ``` -(版本三) 使用defaultdict -```python -from collections import defaultdict - -class Solution: - def isAnagram(self, s: str, t: str) -> bool: - - s_dict = defaultdict(int) - t_dict = defaultdict(int) - - for x in s: - s_dict[x] += 1 - - for x in t: - t_dict[x] += 1 - - return s_dict == t_dict -``` -(版本四) 使用字典 - -```python -class Solution: - def isAnagram(self, s: str, t: str) -> bool: - if len(s) != len(t): - return False - - hash_table_s = {} - hash_table_t = {} - - for i in range(len(s)): - hash_table_s[s[i]] = hash_table_s.get(s[i], 0) + 1 - hash_table_t[t[i]] = hash_table_t.get(t[i], 0) + 1 - - return hash_table_s == hash_table_t -``` - -(版本五) 使用排序 - -```python -class Solution: - def isAnagram(self, s: str, t: str) -> bool: - return sorted(s) == sorted(t) -``` -(版本六) 使用Counter - -```python -from collections import Counter - -class Solution: - def isAnagram(self, s: str, t: str) -> bool: - return Counter(s) == Counter(t) -``` Go： ```go From 6e88b2034d70d261a9cd0388cdcf906e50677195 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E7=A8=8B=E5=BA=8F=E5=91=98Carl?= Date: Sun, 7 May 2023 09:54:15 +0800 Subject: [PATCH 126/154] =?UTF-8?q?Update=200242.=E6=9C=89=E6=95=88?= =?UTF-8?q?=E7=9A=84=E5=AD=97=E6=AF=8D=E5=BC=82=E4=BD=8D=E8=AF=8D.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0242.有效的字母异位词.md | 24 +++++++++++++++++++++++ 1 file changed, 24 insertions(+) diff --git a/problems/0242.有效的字母异位词.md b/problems/0242.有效的字母异位词.md index 3f4b00dc..09674ecd 100644 --- a/problems/0242.有效的字母异位词.md +++ b/problems/0242.有效的字母异位词.md @@ -138,7 +138,31 @@ class Solution: return False return True ``` +Python写法二（没有使用数组作为哈希表，只是介绍defaultdict这样一种解题思路）： +```python +class Solution: + def isAnagram(self, s: str, t: str) -> bool: + from collections import defaultdict + + s_dict = defaultdict(int) + t_dict = defaultdict(int) + for x in s: + s_dict[x] += 1 + + for x in t: + t_dict[x] += 1 + return s_dict == t_dict +``` +Python写法三(没有使用数组作为哈希表，只是介绍Counter这种更方便的解题思路)： + +```python +class Solution(object): + def isAnagram(self, s: str, t: str) -> bool: + from collections import Counter + a_count = Counter(s) + b_count = Counter(t) + return a_count == b_count Go： From d2abce01594ae1bbd73889040c82b35455b3d93e Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Sat, 6 May 2023 22:54:54 -0500 Subject: [PATCH 127/154] =?UTF-8?q?Update=200112.=E8=B7=AF=E5=BE=84?= =?UTF-8?q?=E6=80=BB=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0112.路径总和.md | 267 ++++++++++++++++++++-------------- 1 file changed, 159 insertions(+), 108 deletions(-) diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index 5958de93..9ee2464e 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -451,140 +451,191 @@ class Solution { ### 0112.路径总和 -**递归** - +(版本一) 递归 ```python -class solution: - def haspathsum(self, root: treenode, targetsum: int) -> bool: - def isornot(root, targetsum) -> bool: - if (not root.left) and (not root.right) and targetsum == 0: - return true # 遇到叶子节点，并且计数为0 - if (not root.left) and (not root.right): - return false # 遇到叶子节点，计数不为0 - if root.left: - targetsum -= root.left.val # 左节点 - if isornot(root.left, targetsum): return true # 递归，处理左节点 - targetsum += root.left.val # 回溯 - if root.right: - targetsum -= root.right.val # 右节点 - if isornot(root.right, targetsum): return true # 递归，处理右节点 - targetsum += root.right.val # 回溯 - return false - - if root == none: - return false # 别忘记处理空treenode - else: - return isornot(root, targetsum - root.val) +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def traversal(self, cur: TreeNode, count: int) -> bool: + if not cur.left and not cur.right and count == 0: # 遇到叶子节点，并且计数为0 + return True + if not cur.left and not cur.right: # 遇到叶子节点直接返回 + return False + + if cur.left: # 左 + count -= cur.left.val + if self.traversal(cur.left, count): # 递归，处理节点 + return True + count += cur.left.val # 回溯，撤销处理结果 -class Solution: # 简洁版 - def hasPathSum(self, root: Optional[TreeNode], targetSum: int) -> bool: - if not root: return False - if root.left==root.right==None and root.val == targetSum: return True - return self.hasPathSum(root.left,targetSum-root.val) or self.hasPathSum(root.right,targetSum-root.val) + if cur.right: # 右 + count -= cur.right.val + if self.traversal(cur.right, count): # 递归，处理节点 + return True + count += cur.right.val # 回溯，撤销处理结果 + + return False + + def hasPathSum(self, root: TreeNode, sum: int) -> bool: + if root is None: + return False + return self.traversal(root, sum - root.val) ``` -**迭代 - 层序遍历** - +(版本二) 递归 + 精简 ```python -class solution: - def haspathsum(self, root: treenode, targetsum: int) -> bool: +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def hasPathSum(self, root: TreeNode, sum: int) -> bool: if not root: - return false - - stack = [] # [(当前节点，路径数值), ...] - stack.append((root, root.val)) - - while stack: - cur_node, path_sum = stack.pop() - - if not cur_node.left and not cur_node.right and path_sum == targetsum: - return true - - if cur_node.right: - stack.append((cur_node.right, path_sum + cur_node.right.val)) - - if cur_node.left: - stack.append((cur_node.left, path_sum + cur_node.left.val)) - - return false + return False + if not root.left and not root.right and sum == root.val: + return True + return self.hasPathSum(root.left, sum - root.val) or self.hasPathSum(root.right, sum - root.val) + ``` +(版本三) 迭代 +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def hasPathSum(self, root: TreeNode, sum: int) -> bool: + if not root: + return False + # 此时栈里要放的是pair<节点指针，路径数值> + st = [(root, root.val)] + while st: + node, path_sum = st.pop() + # 如果该节点是叶子节点了，同时该节点的路径数值等于sum，那么就返回true + if not node.left and not node.right and path_sum == sum: + return True + # 右节点，压进去一个节点的时候，将该节点的路径数值也记录下来 + if node.right: + st.append((node.right, path_sum + node.right.val)) + # 左节点，压进去一个节点的时候，将该节点的路径数值也记录下来 + if node.left: + st.append((node.left, path_sum + node.left.val)) + return False + + + +``` + + ### 0113.路径总和-ii -**递归** - +(版本一) 递归 ```python -class solution: - def pathsum(self, root: treenode, targetsum: int) -> list[list[int]]: +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def __init__(self): + self.result = [] + self.path = [] - def traversal(cur_node, remain): - if not cur_node.left and not cur_node.right: - if remain == 0: - result.append(path[:]) - return + def traversal(self, cur, count): + if not cur.left and not cur.right and count == 0: # 遇到了叶子节点且找到了和为sum的路径 + self.result.append(self.path[:]) + return - if cur_node.left: - path.append(cur_node.left.val) - traversal(cur_node.left, remain-cur_node.left.val) - path.pop() + if not cur.left and not cur.right: # 遇到叶子节点而没有找到合适的边，直接返回 + return - if cur_node.right: - path.append(cur_node.right.val) - traversal(cur_node.right, remain-cur_node.right.val) - path.pop() + if cur.left: # 左 （空节点不遍历） + self.path.append(cur.left.val) + count -= cur.left.val + self.traversal(cur.left, count) # 递归 + count += cur.left.val # 回溯 + self.path.pop() # 回溯 - result, path = [], [] + if cur.right: # 右 （空节点不遍历） + self.path.append(cur.right.val) + count -= cur.right.val + self.traversal(cur.right, count) # 递归 + count += cur.right.val # 回溯 + self.path.pop() # 回溯 + + return + + def pathSum(self, root: TreeNode, sum: int) -> List[List[int]]: + self.result.clear() + self.path.clear() if not root: - return [] - path.append(root.val) - traversal(root, targetsum - root.val) - return result + return self.result + self.path.append(root.val) # 把根节点放进路径 + self.traversal(root, sum - root.val) + return self.result ``` -**迭代法，用第二个队列保存目前的总和与路径** - +(版本二) 递归 + 精简 ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: - def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]: - if not root: - return [] - que, temp = deque([root]), deque([(root.val, [root.val])]) + def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]: + result = [] - while que: - for _ in range(len(que)): - node = que.popleft() - value, path = temp.popleft() - if (not node.left) and (not node.right): - if value == targetSum: - result.append(path) - if node.left: - que.append(node.left) - temp.append((node.left.val+value, path+[node.left.val])) - if node.right: - que.append(node.right) - temp.append((node.right.val+value, path+[node.right.val])) + self.traversal(root, targetSum, [], result) return result + def traversal(self,node, count, path, result): + if not node: + return + path.append(node.val) + count -= node.val + if not node.left and not node.right and count == 0: + result.append(list(path)) + self.traversal(node.left, count, path, result) + self.traversal(node.right, count, path, result) + path.pop() ``` - -**迭代法，前序遍历** - +(版本三) 迭代 ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: - def pathSum(self, root: Optional[TreeNode], targetSum: int) -> List[List[int]]: - if not root: return [] - stack, path_stack,result = [[root,root.val]],[[root.val]],[] + def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]: + if not root: + return [] + stack = [(root, [root.val])] + res = [] while stack: - cur,cursum = stack.pop() - path = path_stack.pop() - if cur.left==cur.right==None: - if cursum==targetSum: result.append(path) - if cur.right: - stack.append([cur.right,cursum+cur.right.val]) - path_stack.append(path+[cur.right.val]) - if cur.left: - stack.append([cur.left,cursum+cur.left.val]) - path_stack.append(path+[cur.left.val]) - return result + node, path = stack.pop() + if not node.left and not node.right and sum(path) == targetSum: + res.append(path) + if node.right: + stack.append((node.right, path + [node.right.val])) + if node.left: + stack.append((node.left, path + [node.left.val])) + return res + + + ``` ## go From 40f8230fd86b475e87d970f973042a2b4cbee613 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 9 May 2023 00:20:58 -0500 Subject: [PATCH 128/154] =?UTF-8?q?Update=200106.=E4=BB=8E=E4=B8=AD?= =?UTF-8?q?=E5=BA=8F=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86=E5=BA=8F?= =?UTF-8?q?=E5=88=97=E6=9E=84=E9=80=A0=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit python 代码重复 --- ...序与后序遍历序列构造二叉树.md | 38 +------------------ 1 file changed, 2 insertions(+), 36 deletions(-) diff --git a/problems/0106.从中序与后序遍历序列构造二叉树.md b/problems/0106.从中序与后序遍历序列构造二叉树.md index adb374f9..89afad08 100644 --- a/problems/0106.从中序与后序遍历序列构造二叉树.md +++ b/problems/0106.从中序与后序遍历序列构造二叉树.md @@ -400,8 +400,6 @@ public: }; ``` -## Python - # 105.从前序与中序遍历序列构造二叉树 @@ -657,38 +655,6 @@ class Solution { ## Python -```python -class Solution: - def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]: - # 第一步: 特殊情况讨论: 树为空. 或者说是递归终止条件 - if not postorder: - return - - # 第二步: 后序遍历的最后一个就是当前的中间节点 - root_val = postorder[-1] - root = TreeNode(root_val) - - # 第三步: 找切割点. - root_index = inorder.index(root_val) - - # 第四步: 切割inorder数组. 得到inorder数组的左,右半边. - left_inorder = inorder[:root_index] - right_inorder = inorder[root_index + 1:] - - # 第五步: 切割postorder数组. 得到postorder数组的左,右半边. - # ⭐️ 重点1: 中序数组大小一定跟后序数组大小是相同的. - left_postorder = postorder[:len(left_inorder)] - right_postorder = postorder[len(left_inorder): len(postorder) - 1] - - - # 第六步: 递归 - root.left = self.buildTree(left_inorder, left_postorder) - root.right = self.buildTree(right_inorder, right_postorder) - - # 第七步: 返回答案 - return root -``` - 105.从前序与中序遍历序列构造二叉树 ```python @@ -717,7 +683,7 @@ class Solution: # 第六步: 递归 root.left = self.buildTree(preorder_left, inorder_left) root.right = self.buildTree(preorder_right, inorder_right) - + # 第七步: 返回答案 return root ``` @@ -749,7 +715,7 @@ class Solution: # 第六步: 递归 root.left = self.buildTree(inorder_left, postorder_left) root.right = self.buildTree(inorder_right, postorder_right) - + # 第七步: 返回答案 return root ``` From 7eea41ee0e18381088bce52f7ea218e1ffc78c77 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 9 May 2023 14:04:22 -0500 Subject: [PATCH 129/154] =?UTF-8?q?Update=200654.=E6=9C=80=E5=A4=A7?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0654.最大二叉树.md | 92 +++++++++++++++++++------------- 1 file changed, 55 insertions(+), 37 deletions(-) diff --git a/problems/0654.最大二叉树.md b/problems/0654.最大二叉树.md index 64b38b48..224e75e4 100644 --- a/problems/0654.最大二叉树.md +++ b/problems/0654.最大二叉树.md @@ -259,52 +259,70 @@ class Solution { ``` ### Python - +(版本一) 基础版 ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: - """递归法 更快""" def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode: - if not nums: - return None - maxvalue = max(nums) - index = nums.index(maxvalue) - - root = TreeNode(maxvalue) - - left = nums[:index] - right = nums[index + 1:] - - root.left = self.constructMaximumBinaryTree(left) - root.right = self.constructMaximumBinaryTree(right) - return root + if len(nums) == 1: + return TreeNode(nums[0]) + node = TreeNode(0) + # 找到数组中最大的值和对应的下标 + maxValue = 0 + maxValueIndex = 0 + for i in range(len(nums)): + if nums[i] > maxValue: + maxValue = nums[i] + maxValueIndex = i + node.val = maxValue + # 最大值所在的下标左区间 构造左子树 + if maxValueIndex > 0: + new_list = nums[:maxValueIndex] + node.left = self.constructMaximumBinaryTree(new_list) + # 最大值所在的下标右区间 构造右子树 + if maxValueIndex < len(nums) - 1: + new_list = nums[maxValueIndex+1:] + node.right = self.constructMaximumBinaryTree(new_list) + return node +``` +(版本二) 使用下标 class Solution: - """最大二叉树 递归法""" + def traversal(self, nums: List[int], left: int, right: int) -> TreeNode: + if left >= right: + return None + maxValueIndex = left + for i in range(left + 1, right): + if nums[i] > nums[maxValueIndex]: + maxValueIndex = i + root = TreeNode(nums[maxValueIndex]) + root.left = self.traversal(nums, left, maxValueIndex) + root.right = self.traversal(nums, maxValueIndex + 1, right) + return root def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode: return self.traversal(nums, 0, len(nums)) - - def traversal(self, nums: List[int], begin: int, end: int) -> TreeNode: - # 列表长度为0时返回空节点 - if begin == end: - return None - - # 找到最大的值和其对应的下标 - max_index = begin - for i in range(begin, end): - if nums[i] > nums[max_index]: - max_index = i - - # 构建当前节点 - root = TreeNode(nums[max_index]) - - # 递归构建左右子树 - root.left = self.traversal(nums, begin, max_index) - root.right = self.traversal(nums, max_index + 1, end) - - return root -``` + +``` +(版本三) 使用切片 +class Solution: + def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode: + if not nums: + return None + max_val = max(nums) + max_index = nums.index(max_val) + node = TreeNode(max_val) + node.left = self.constructMaximumBinaryTree(nums[:max_index]) + node.right = self.constructMaximumBinaryTree(nums[max_index+1:]) + return node + +``` ### Go From a45046be8a6387a0c4e7fbf6872c554b6a2f3249 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 9 May 2023 15:11:13 -0500 Subject: [PATCH 130/154] =?UTF-8?q?Update=200617.=E5=90=88=E5=B9=B6?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加 创建 root节点 和 python queue的代码优化版本 --- problems/0617.合并二叉树.md | 67 ++++++++++++++++++++++++++++++-- 1 file changed, 64 insertions(+), 3 deletions(-) diff --git a/problems/0617.合并二叉树.md b/problems/0617.合并二叉树.md index f98948f0..18a245c4 100644 --- a/problems/0617.合并二叉树.md +++ b/problems/0617.合并二叉树.md @@ -352,8 +352,7 @@ class Solution { ``` ### Python - -**递归法 - 前序遍历** +(版本一) 递归 - 前序 - 修改root1 ```python # Definition for a binary tree node. # class TreeNode: @@ -377,8 +376,33 @@ class Solution: return root1 # ⚠️ 注意: 本题我们重复使用了题目给出的节点而不是创建新节点. 节省时间, 空间. ``` +(版本二) 递归 - 前序 - 新建root +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode: + # 递归终止条件: + # 但凡有一个节点为空, 就立刻返回另外一个. 如果另外一个也为None就直接返回None. + if not root1: + return root2 + if not root2: + return root1 + # 上面的递归终止条件保证了代码执行到这里root1, root2都非空. + root = TreeNode() # 创建新节点 + root.val += root1.val + root2.val# 中 + root.left = self.mergeTrees(root1.left, root2.left) #左 + root.right = self.mergeTrees(root1.right, root2.right) # 右 + + return root # ⚠️ 注意: 本题我们创建了新节点. -**迭代法** +``` + +(版本三) 迭代 ```python class Solution: def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode: @@ -413,7 +437,44 @@ class Solution: return root1 ``` +(版本四) 迭代 + 代码优化 +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +from collections import deque +class Solution: + def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode: + if not root1: + return root2 + if not root2: + return root1 + + queue = deque() + queue.append((root1, root2)) + + while queue: + node1, node2 = queue.popleft() + node1.val += node2.val + + if node1.left and node2.left: + queue.append((node1.left, node2.left)) + elif not node1.left: + node1.left = node2.left + + if node1.right and node2.right: + queue.append((node1.right, node2.right)) + elif not node1.right: + node1.right = node2.right + + return root1 + + +``` ### Go ```go From 3818de4610acd3a1a3e1a9794f3e6caa4320a9e9 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 9 May 2023 18:50:14 -0500 Subject: [PATCH 131/154] =?UTF-8?q?Update=200700.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84=E6=90=9C=E7=B4=A2?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0700.二叉搜索树中的搜索.md | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/problems/0700.二叉搜索树中的搜索.md b/problems/0700.二叉搜索树中的搜索.md index 40724f48..13064f97 100644 --- a/problems/0700.二叉搜索树中的搜索.md +++ b/problems/0700.二叉搜索树中的搜索.md @@ -230,7 +230,7 @@ class Solution { ### Python -递归法： +（方法一） 递归 ```python class Solution: @@ -250,12 +250,12 @@ class Solution: ``` -迭代法： +（方法二）迭代 ```python class Solution: def searchBST(self, root: TreeNode, val: int) -> TreeNode: - while root is not None: + while root: if val < root.val: root = root.left elif val > root.val: root = root.right else: return root From 16f4a48bd6b0539c60eed95c4f2c54a052685386 Mon Sep 17 00:00:00 2001 From: programmercarl <826123027@qq.com> Date: Sat, 20 May 2023 15:30:25 +0800 Subject: [PATCH 132/154] update --- problems/0053.最大子序和.md | 2 +- problems/0121.买卖股票的最佳时机.md | 17 +++++++----- .../0122.买卖股票的最佳时机II.md | 2 +- ...票的最佳时机II（动态规划）.md | 23 +++++++++------- .../0123.买卖股票的最佳时机III.md | 27 +++++++++++-------- .../0188.买卖股票的最佳时机IV.md | 17 +++++++----- problems/0198.打家劫舍.md | 4 +++ problems/0213.打家劫舍II.md | 25 ++++++++++------- ...09.最佳买卖股票时机含冷冻期.md | 4 +++ problems/0337.打家劫舍III.md | 5 ++++ ...佳时机含手续费（动态规划）.md | 5 ++++ ...1005.K次取反后最大化的数组和.md | 23 ---------------- problems/动态规划理论基础.md | 5 ++++ 13 files changed, 92 insertions(+), 67 deletions(-) diff --git a/problems/0053.最大子序和.md b/problems/0053.最大子序和.md index 48f8be29..39c58332 100644 --- a/problems/0053.最大子序和.md +++ b/problems/0053.最大子序和.md @@ -124,7 +124,7 @@ public: ## 动态规划 -当然本题还可以用动态规划来做，当前[「代码随想录」](https://img-blog.csdnimg.cn/20201124161234338.png)主要讲解贪心系列，后续到动态规划系列的时候会详细讲解本题的 dp 方法。 +当然本题还可以用动态规划来做，在代码随想录动态规划章节我会详细介绍，如果大家想在想看，可以直接跳转：[动态规划版本详解](https://programmercarl.com/0053.%E6%9C%80%E5%A4%A7%E5%AD%90%E5%BA%8F%E5%92%8C%EF%BC%88%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%EF%BC%89.html#%E6%80%9D%E8%B7%AF) 那么先给出我的 dp 代码如下，有时间的录友可以提前做一做： diff --git a/problems/0121.买卖股票的最佳时机.md b/problems/0121.买卖股票的最佳时机.md index 5b398f3f..753cb106 100644 --- a/problems/0121.买卖股票的最佳时机.md +++ b/problems/0121.买卖股票的最佳时机.md @@ -14,16 +14,21 @@ 返回你可以从这笔交易中获取的最大利润。如果你不能获取任何利润，返回 0 。 -示例 1： -输入：[7,1,5,3,6,4] -输出：5 +* 示例 1： +* 输入：[7,1,5,3,6,4] +* 输出：5 解释：在第 2 天（股票价格 = 1）的时候买入，在第 5 天（股票价格 = 6）的时候卖出，最大利润 = 6-1 = 5 。注意利润不能是 7-1 = 6, 因为卖出价格需要大于买入价格；同时，你不能在买入前卖出股票。 -示例 2： -输入：prices = [7,6,4,3,1] -输出：0 +* 示例 2： +* 输入：prices = [7,6,4,3,1] +* 输出：0 解释：在这种情况下, 没有交易完成, 所以最大利润为 0。 +# 算法公开课 + +**《代码随想录》算法视频公开课：[动态规划之 LeetCode：121.买卖股票的最佳时机1](https://www.bilibili.com/video/BV1Xe4y1u77q)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 + + ## 思路 diff --git a/problems/0122.买卖股票的最佳时机II.md b/problems/0122.买卖股票的最佳时机II.md index e631c5e2..0d8ad608 100644 --- a/problems/0122.买卖股票的最佳时机II.md +++ b/problems/0122.买卖股票的最佳时机II.md @@ -102,7 +102,7 @@ public: ### 动态规划 -动态规划将在下一个系列详细讲解，本题解先给出我的 C++代码（带详细注释），感兴趣的同学可以自己先学习一下。 +动态规划将在下一个系列详细讲解，本题解先给出我的 C++代码（带详细注释），想先学习的话，可以看本篇：[122.买卖股票的最佳时机II（动态规划）](https://programmercarl.com/0122.%E4%B9%B0%E5%8D%96%E8%82%A1%E7%A5%A8%E7%9A%84%E6%9C%80%E4%BD%B3%E6%97%B6%E6%9C%BAII%EF%BC%88%E5%8A%A8%E6%80%81%E8%A7%84%E5%88%92%EF%BC%89.html#%E6%80%9D%E8%B7%AF) ```CPP class Solution { diff --git a/problems/0122.买卖股票的最佳时机II（动态规划）.md b/problems/0122.买卖股票的最佳时机II（动态规划）.md index 2779083d..146c6a4c 100644 --- a/problems/0122.买卖股票的最佳时机II（动态规划）.md +++ b/problems/0122.买卖股票的最佳时机II（动态规划）.md @@ -15,25 +15,30 @@ 注意：你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。 -示例 1: -输入: [7,1,5,3,6,4] -输出: 7 +* 示例 1: +* 输入: [7,1,5,3,6,4] +* 输出: 7 解释: 在第 2 天（股票价格 = 1）的时候买入，在第 3 天（股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5-1 = 4。随后，在第 4 天（股票价格 = 3）的时候买入，在第 5 天（股票价格 = 6）的时候卖出, 这笔交易所能获得利润 = 6-3 = 3 。 -示例 2: -输入: [1,2,3,4,5] -输出: 4 +* 示例 2: +* 输入: [1,2,3,4,5] +* 输出: 4 解释: 在第 1 天（股票价格 = 1）的时候买入，在第 5 天 （股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5-1 = 4 。注意你不能在第 1 天和第 2 天接连购买股票，之后再将它们卖出。因为这样属于同时参与了多笔交易，你必须在再次购买前出售掉之前的股票。 -示例 3: -输入: [7,6,4,3,1] -输出: 0 +* 示例 3: +* 输入: [7,6,4,3,1] +* 输出: 0 解释: 在这种情况下, 没有交易完成, 所以最大利润为 0。 提示： * 1 <= prices.length <= 3 * 10 ^ 4 * 0 <= prices[i] <= 10 ^ 4 +# 算法公开课 + +**《代码随想录》算法视频公开课：[动态规划，股票问题第二弹 | LeetCode：122.买卖股票的最佳时机II](https://www.bilibili.com/video/BV1D24y1Q7Ls)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 + + ## 思路 本题我们在讲解贪心专题的时候就已经讲解过了[贪心算法：买卖股票的最佳时机II](https://programmercarl.com/0122.买卖股票的最佳时机II.html)，只不过没有深入讲解动态规划的解法，那么这次我们再好好分析一下动规的解法。 diff --git a/problems/0123.买卖股票的最佳时机III.md b/problems/0123.买卖股票的最佳时机III.md index 33157238..af6870d4 100644 --- a/problems/0123.买卖股票的最佳时机III.md +++ b/problems/0123.买卖股票的最佳时机III.md @@ -15,23 +15,23 @@ 注意：你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。 -示例 1: -输入：prices = [3,3,5,0,0,3,1,4] -输出：6 +* 示例 1: +* 输入：prices = [3,3,5,0,0,3,1,4] +* 输出：6 解释：在第 4 天（股票价格 = 0）的时候买入，在第 6 天（股票价格 = 3）的时候卖出，这笔交易所能获得利润 = 3-0 = 3 。随后，在第 7 天（股票价格 = 1）的时候买入，在第 8 天 （股票价格 = 4）的时候卖出，这笔交易所能获得利润 = 4-1 = 3。 -示例 2： -输入：prices = [1,2,3,4,5] -输出：4 +* 示例 2： +* 输入：prices = [1,2,3,4,5] +* 输出：4 解释：在第 1 天（股票价格 = 1）的时候买入，在第 5 天 （股票价格 = 5）的时候卖出, 这笔交易所能获得利润 = 5-1 = 4。注意你不能在第 1 天和第 2 天接连购买股票，之后再将它们卖出。因为这样属于同时参与了多笔交易，你必须在再次购买前出售掉之前的股票。 -示例 3： -输入：prices = [7,6,4,3,1] -输出：0 +* 示例 3： +* 输入：prices = [7,6,4,3,1] +* 输出：0 解释：在这个情况下, 没有交易完成, 所以最大利润为0。 -示例 4： -输入：prices = [1] +* 示例 4： +* 输入：prices = [1] 输出：0 提示： @@ -39,6 +39,11 @@ * 1 <= prices.length <= 10^5 * 0 <= prices[i] <= 10^5 +# 算法公开课 + +**《代码随想录》算法视频公开课：[动态规划，股票至多买卖两次，怎么求？ | LeetCode：123.买卖股票最佳时机III](https://www.bilibili.com/video/BV1WG411K7AR)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 + + ## 思路 diff --git a/problems/0188.买卖股票的最佳时机IV.md b/problems/0188.买卖股票的最佳时机IV.md index f6744a2b..37270664 100644 --- a/problems/0188.买卖股票的最佳时机IV.md +++ b/problems/0188.买卖股票的最佳时机IV.md @@ -14,14 +14,14 @@ 注意：你不能同时参与多笔交易（你必须在再次购买前出售掉之前的股票）。 -示例 1： -输入：k = 2, prices = [2,4,1] -输出：2 +* 示例 1： +* 输入：k = 2, prices = [2,4,1] +* 输出：2 解释：在第 1 天 (股票价格 = 2) 的时候买入，在第 2 天 (股票价格 = 4) 的时候卖出，这笔交易所能获得利润 = 4-2 = 2。 -示例 2： -输入：k = 2, prices = [3,2,6,5,0,3] -输出：7 +* 示例 2： +* 输入：k = 2, prices = [3,2,6,5,0,3] +* 输出：7 解释：在第 2 天 (股票价格 = 2) 的时候买入，在第 3 天 (股票价格 = 6) 的时候卖出, 这笔交易所能获得利润 = 6-2 = 4。随后，在第 5 天 (股票价格 = 0) 的时候买入，在第 6 天 (股票价格 = 3) 的时候卖出, 这笔交易所能获得利润 = 3-0 = 3 。 @@ -31,6 +31,11 @@ * 0 <= prices.length <= 1000 * 0 <= prices[i] <= 1000 +# 算法公开课 + +**《代码随想录》算法视频公开课：[动态规划来决定最佳时机，至多可以买卖K次！| LeetCode：188.买卖股票最佳时机4](https://www.bilibili.com/video/BV16M411U7XJ)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 + + ## 思路 这道题目可以说是[动态规划：123.买卖股票的最佳时机III](https://programmercarl.com/0123.买卖股票的最佳时机III.html)的进阶版，这里要求至多有k次交易。 diff --git a/problems/0198.打家劫舍.md b/problems/0198.打家劫舍.md index c25f3b86..b7cdc1ce 100644 --- a/problems/0198.打家劫舍.md +++ b/problems/0198.打家劫舍.md @@ -31,6 +31,10 @@ * 0 <= nums.length <= 100 * 0 <= nums[i] <= 400 +# 算法公开课 + +**《代码随想录》算法视频公开课：[动态规划，偷不偷这个房间呢？| LeetCode：198.打家劫舍](https://www.bilibili.com/video/BV1Te411N7SX)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 + ## 思路 diff --git a/problems/0213.打家劫舍II.md b/problems/0213.打家劫舍II.md index becad069..aebda7f5 100644 --- a/problems/0213.打家劫舍II.md +++ b/problems/0213.打家劫舍II.md @@ -14,23 +14,28 @@ 示例 1： -输入：nums = [2,3,2] -输出：3 -解释：你不能先偷窃 1 号房屋（金额 = 2），然后偷窃 3 号房屋（金额 = 2）, 因为他们是相邻的。 +* 输入：nums = [2,3,2] +* 输出：3 +* 解释：你不能先偷窃 1 号房屋（金额 = 2），然后偷窃 3 号房屋（金额 = 2）, 因为他们是相邻的。 -示例 2： -输入：nums = [1,2,3,1] -输出：4 -解释：你可以先偷窃 1 号房屋（金额 = 1），然后偷窃 3 号房屋（金额 = 3）。偷窃到的最高金额 = 1 + 3 = 4 。 +* 示例 2： +* 输入：nums = [1,2,3,1] +* 输出：4 +* 解释：你可以先偷窃 1 号房屋（金额 = 1），然后偷窃 3 号房屋（金额 = 3）。偷窃到的最高金额 = 1 + 3 = 4 。 -示例 3： -输入：nums = [0] -输出：0 +* 示例 3： +* 输入：nums = [0] +* 输出：0 提示： * 1 <= nums.length <= 100 * 0 <= nums[i] <= 1000 +# 算法公开课 + +**《代码随想录》算法视频公开课：[动态规划，房间连成环了那还偷不偷呢？| LeetCode：213.打家劫舍II](https://www.bilibili.com/video/BV1oM411B7xq)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 + + ## 思路 这道题目和[198.打家劫舍](https://programmercarl.com/0198.打家劫舍.html)是差不多的，唯一区别就是成环了。 diff --git a/problems/0309.最佳买卖股票时机含冷冻期.md b/problems/0309.最佳买卖股票时机含冷冻期.md index d62e91c7..a56d9b84 100644 --- a/problems/0309.最佳买卖股票时机含冷冻期.md +++ b/problems/0309.最佳买卖股票时机含冷冻期.md @@ -20,6 +20,10 @@ * 输出: 3 * 解释: 对应的交易状态为: [买入, 卖出, 冷冻期, 买入, 卖出] +# 算法公开课 + +**《代码随想录》算法视频公开课：[动态规划来决定最佳时机，这次有冷冻期！| LeetCode：309.买卖股票的最佳时机含冷冻期](https://www.bilibili.com/video/BV1rP4y1D7ku)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 + ## 思路 diff --git a/problems/0337.打家劫舍III.md b/problems/0337.打家劫舍III.md index f75c4c88..ca8cea23 100644 --- a/problems/0337.打家劫舍III.md +++ b/problems/0337.打家劫舍III.md @@ -16,6 +16,11 @@ ![337.打家劫舍III](https://code-thinking-1253855093.file.myqcloud.com/pics/20210223173849619.png) +# 算法公开课 + +**《代码随想录》算法视频公开课：[动态规划，房间连成树了，偷不偷呢？| LeetCode：337.打家劫舍3](https://www.bilibili.com/video/BV1H24y1Q7sY)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 + + ## 思路 这道题目和 [198.打家劫舍](https://programmercarl.com/0198.打家劫舍.html)，[213.打家劫舍II](https://programmercarl.com/0213.打家劫舍II.html)也是如出一辙，只不过这个换成了树。 diff --git a/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md b/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md index 39730807..12789934 100644 --- a/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md +++ b/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md @@ -32,6 +32,11 @@ * 0 < prices[i] < 50000. * 0 <= fee < 50000. +# 算法公开课 + +**《代码随想录》算法视频公开课：[动态规划来决定最佳时机，这次含手续费！| LeetCode：714.买卖股票的最佳时机含手续费](https://www.bilibili.com/video/BV1z44y1Z7UR)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 + + ## 思路 本题贪心解法：[贪心算法：买卖股票的最佳时机含手续费](https://programmercarl.com/0714.买卖股票的最佳时机含手续费.html) diff --git a/problems/1005.K次取反后最大化的数组和.md b/problems/1005.K次取反后最大化的数组和.md index 27a575c7..6252c697 100644 --- a/problems/1005.K次取反后最大化的数组和.md +++ b/problems/1005.K次取反后最大化的数组和.md @@ -130,29 +130,6 @@ class Solution { } ``` -```java -class Solution { - public int largestSumAfterKNegations(int[] A, int K) { - if (A.length == 1) return k % 2 == 0 ? A[0] : -A[0]; - Arrays.sort(A); - int sum = 0; - int idx = 0; - for (int i = 0; i < K; i++) { - if (i < A.length - 1 && A[idx] < 0) { - A[idx] = -A[idx]; - if (A[idx] >= Math.abs(A[idx + 1])) idx++; - continue; - } - A[idx] = -A[idx]; - } - - for (int i = 0; i < A.length; i++) { - sum += A[i]; - } - return sum; - } -} -``` ### Python ```python diff --git a/problems/动态规划理论基础.md b/problems/动态规划理论基础.md index 77f01b13..a99c0690 100644 --- a/problems/动态规划理论基础.md +++ b/problems/动态规划理论基础.md @@ -10,6 +10,11 @@ +## 算法公开课 + +**《代码随想录》算法视频公开课：[动态规划理论基础](https://www.bilibili.com/video/BV13Q4y197Wg)，相信结合视频再看本篇题解，更有助于大家对本题的理解**。 + + ## 什么是动态规划 动态规划，英文：Dynamic Programming，简称DP，如果某一问题有很多重叠子问题，使用动态规划是最有效的。 From f5329cd4b913d73f049e6b30f7d2a3a8a2a90770 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 22 May 2023 15:38:38 -0500 Subject: [PATCH 133/154] =?UTF-8?q?Update=200110.=E5=B9=B3=E8=A1=A1?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 48 ++++++++++++++++++++++++++++++-- 1 file changed, 46 insertions(+), 2 deletions(-) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index a3bc77fb..c0561e10 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -553,6 +553,52 @@ class Solution: 迭代法： +```python +class Solution: + def getDepth(self, cur): + st = [] + if cur is not None: + st.append(cur) + depth = 0 + result = 0 + while st: + node = st[-1] + if node is not None: + st.pop() + st.append(node) # 中 + st.append(None) + depth += 1 + if node.right: + st.append(node.right) # 右 + if node.left: + st.append(node.left) # 左 + + else: + node = st.pop() + st.pop() + depth -= 1 + result = max(result, depth) + return result + + def isBalanced(self, root): + st = [] + if root is None: + return True + st.append(root) + while st: + node = st.pop() # 中 + if abs(self.getDepth(node.left) - self.getDepth(node.right)) > 1: + return False + if node.right: + st.append(node.right) # 右（空节点不入栈） + if node.left: + st.append(node.left) # 左（空节点不入栈） + return True + +``` + +迭代法精简版： + ```python class Solution: def isBalanced(self, root: Optional[TreeNode]) -> bool: @@ -576,8 +622,6 @@ class Solution: height_map[real_node] = 1 + max(left, right) return True ``` - - ### Go ```Go From 2024fc272af5811c3c594bf08959af2c9f62371d Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 22 May 2023 16:48:27 -0500 Subject: [PATCH 134/154] =?UTF-8?q?Update=200257.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=89=80=E6=9C=89=E8=B7=AF=E5=BE=84.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0257.二叉树的所有路径.md | 40 ++++++++++------------- 1 file changed, 18 insertions(+), 22 deletions(-) diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md index c396f4a0..4cafbded 100644 --- a/problems/0257.二叉树的所有路径.md +++ b/problems/0257.二叉树的所有路径.md @@ -473,31 +473,27 @@ class Solution { 递归法+回溯（版本一） ```Python # Definition for a binary tree node. -# class TreeNode: -# def __init__(self, val=0, left=None, right=None): -# self.val = val -# self.left = left -# self.right = right -import copy -from typing import List, Optional - class Solution: - def binaryTreePaths(self, root: Optional[TreeNode]) -> List[str]: - if not root: - return [] + def traversal(self, cur, path, result): + path.append(cur.val) # 中 + if not cur.left and not cur.right: # 到达叶子节点 + sPath = '->'.join(map(str, path)) + result.append(sPath) + return + if cur.left: # 左 + self.traversal(cur.left, path, result) + path.pop() # 回溯 + if cur.right: # 右 + self.traversal(cur.right, path, result) + path.pop() # 回溯 + + def binaryTreePaths(self, root): result = [] - self.generate_paths(root, [], result) + path = [] + if not root: + return result + self.traversal(root, path, result) return result - - def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None: - path.append(node.val) - if not node.left and not node.right: - result.append('->'.join(map(str, path))) - if node.left: - self.generate_paths(node.left, copy.copy(path), result) - if node.right: - self.generate_paths(node.right, copy.copy(path), result) - path.pop() ``` From 9b770de8cccfee4588a4966086aa4f285c347d68 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 22 May 2023 17:07:10 -0500 Subject: [PATCH 135/154] =?UTF-8?q?Update=200257.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=89=80=E6=9C=89=E8=B7=AF=E5=BE=84.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0257.二叉树的所有路径.md | 11 +++++------ 1 file changed, 5 insertions(+), 6 deletions(-) diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md index 4cafbded..195b6f94 100644 --- a/problems/0257.二叉树的所有路径.md +++ b/problems/0257.二叉树的所有路径.md @@ -497,7 +497,7 @@ class Solution: ``` -递归法+回溯（版本二） +递归法+隐形回溯（版本一） ```Python # Definition for a binary tree node. # class TreeNode: @@ -505,7 +505,6 @@ class Solution: # self.val = val # self.left = left # self.right = right -import copy from typing import List, Optional class Solution: @@ -523,13 +522,13 @@ class Solution: if not node.left and not node.right: result.append('->'.join(map(str, path))) else: - self.generate_paths(node.left, copy.copy(path), result) - self.generate_paths(node.right, copy.copy(path), result) - path.pop() + # path[:] 是隐藏回溯 + self.generate_paths(node.left, path[:], result) + self.generate_paths(node.right, path[:], result) ``` -递归法+隐形回溯 +递归法+隐形回溯（版本二） ```Python # Definition for a binary tree node. # class TreeNode: From 1a672740e99c000e44e1c5186bc57184b1f35b58 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 22 May 2023 17:14:09 -0500 Subject: [PATCH 136/154] =?UTF-8?q?Update=200257.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=89=80=E6=9C=89=E8=B7=AF=E5=BE=84.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0257.二叉树的所有路径.md | 18 +++++++++--------- 1 file changed, 9 insertions(+), 9 deletions(-) diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md index 195b6f94..31cbdb56 100644 --- a/problems/0257.二叉树的所有路径.md +++ b/problems/0257.二叉树的所有路径.md @@ -512,19 +512,19 @@ class Solution: if not root: return [] result = [] - self.generate_paths(root, [], result) + self.traversal(root, [], result) return result - def generate_paths(self, node: TreeNode, path: List[int], result: List[str]) -> None: - if not node: + def traversal(self, cur: TreeNode, path: List[int], result: List[str]) -> None: + if not cur: return - path.append(node.val) - if not node.left and not node.right: + path.append(cur.val) + if not cur.left and not cur.right: result.append('->'.join(map(str, path))) - else: - # path[:] 是隐藏回溯 - self.generate_paths(node.left, path[:], result) - self.generate_paths(node.right, path[:], result) + if cur.left: + self.traversal(cur.left, path[:], result) + if cur.right: + self.traversal(cur.right, path[:], result) ``` From 2e81b18c5d92d99e7b50c35f0db76f8b824b3915 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 22 May 2023 19:04:11 -0500 Subject: [PATCH 137/154] =?UTF-8?q?Update=200257.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=89=80=E6=9C=89=E8=B7=AF=E5=BE=84.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0257.二叉树的所有路径.md | 9 ++------- 1 file changed, 2 insertions(+), 7 deletions(-) diff --git a/problems/0257.二叉树的所有路径.md b/problems/0257.二叉树的所有路径.md index 31cbdb56..7bd56fbd 100644 --- a/problems/0257.二叉树的所有路径.md +++ b/problems/0257.二叉树的所有路径.md @@ -470,7 +470,7 @@ class Solution { ## Python: -递归法+回溯（版本一） +递归法+回溯 ```Python # Definition for a binary tree node. class Solution: @@ -561,16 +561,11 @@ class Solution: 迭代法： ```Python -from collections import deque - - class Solution: - """二叉树的所有路径 迭代法""" def binaryTreePaths(self, root: TreeNode) -> List[str]: # 题目中节点数至少为1 - stack, path_st, result = deque([root]), deque(), [] - path_st.append(str(root.val)) + stack, path_st, result = [root], [str(root.val)], [] while stack: cur = stack.pop() From 117ef697fac9428f8982a9076a9f25445f1d72ee Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 22 May 2023 19:24:48 -0500 Subject: [PATCH 138/154] =?UTF-8?q?Update=200404.=E5=B7=A6=E5=8F=B6?= =?UTF-8?q?=E5=AD=90=E4=B9=8B=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0404.左叶子之和.md | 88 +++++++++++++++++++------------- 1 file changed, 52 insertions(+), 36 deletions(-) diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md index 617978b7..aa2868df 100644 --- a/problems/0404.左叶子之和.md +++ b/problems/0404.左叶子之和.md @@ -247,8 +247,7 @@ class Solution { ### Python - -**递归后序遍历** +递归 ```python # Definition for a binary tree node. # class TreeNode: @@ -257,47 +256,64 @@ class Solution { # self.left = left # self.right = right class Solution: - def sumOfLeftLeaves(self, root: Optional[TreeNode]) -> int: - if not root: + def sumOfLeftLeaves(self, root): + if root is None: + return 0 + if root.left is None and root.right is None: return 0 - # 检查根节点的左子节点是否为叶节点 - if root.left and not root.left.left and not root.left.right: - left_val = root.left.val - else: - left_val = self.sumOfLeftLeaves(root.left) + leftValue = self.sumOfLeftLeaves(root.left) # 左 + if root.left and not root.left.left and not root.left.right: # 左子树是左叶子的情况 + leftValue = root.left.val - # 递归地计算右子树左叶节点的和 - right_val = self.sumOfLeftLeaves(root.right) - - return left_val + right_val + rightValue = self.sumOfLeftLeaves(root.right) # 右 + + sum_val = leftValue + rightValue # 中 + return sum_val +``` +递归精简版 + +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def sumOfLeftLeaves(self, root): + if root is None: + return 0 + leftValue = 0 + if root.left is not None and root.left.left is None and root.left.right is None: + leftValue = root.left.val + return leftValue + self.sumOfLeftLeaves(root.left) + self.sumOfLeftLeaves(root.right) ``` -**迭代** +迭代法 ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: - def sumOfLeftLeaves(self, root: TreeNode) -> int: - """ - Idea: Each time check current node's left node. - If current node don't have one, skip it. - """ - stack = [] - if root: - stack.append(root) - res = 0 - - while stack: - # 每次都把当前节点的左节点加进去. - cur_node = stack.pop() - if cur_node.left and not cur_node.left.left and not cur_node.left.right: - res += cur_node.left.val - - if cur_node.left: - stack.append(cur_node.left) - if cur_node.right: - stack.append(cur_node.right) - - return res + def sumOfLeftLeaves(self, root): + if root is None: + return 0 + st = [root] + result = 0 + while st: + node = st.pop() + if node.left and node.left.left is None and node.left.right is None: + result += node.left.val + if node.right: + st.append(node.right) + if node.left: + st.append(node.left) + return result + ``` ### Go From 1ac1a8c332fde5f37e848577fae4970e46c1d7e9 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 22 May 2023 19:42:19 -0500 Subject: [PATCH 139/154] =?UTF-8?q?Update=200513.=E6=89=BE=E6=A0=91?= =?UTF-8?q?=E5=B7=A6=E4=B8=8B=E8=A7=92=E7=9A=84=E5=80=BC.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0513.找树左下角的值.md | 15 ++++++++++----- 1 file changed, 10 insertions(+), 5 deletions(-) diff --git a/problems/0513.找树左下角的值.md b/problems/0513.找树左下角的值.md index 26768c74..743b0df9 100644 --- a/problems/0513.找树左下角的值.md +++ b/problems/0513.找树左下角的值.md @@ -330,19 +330,24 @@ class Solution: # self.right = right from collections import deque class Solution: - def findBottomLeftValue(self, root: Optional[TreeNode]) -> int: - queue = deque([root]) + def findBottomLeftValue(self, root): + if root is None: + return 0 + queue = deque() + queue.append(root) + result = 0 while queue: size = len(queue) - leftmost = queue[0].val for i in range(size): node = queue.popleft() + if i == 0: + result = node.val if node.left: queue.append(node.left) if node.right: queue.append(node.right) - if not queue: - return leftmost + return result + ``` From 12634b2d89a00e25f2c3897a675c58922c93d0b0 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 22 May 2023 21:48:07 -0500 Subject: [PATCH 140/154] =?UTF-8?q?Update=200654.=E6=9C=80=E5=A4=A7?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0654.最大二叉树.md | 5 ++++- 1 file changed, 4 insertions(+), 1 deletion(-) diff --git a/problems/0654.最大二叉树.md b/problems/0654.最大二叉树.md index 224e75e4..07fbeb82 100644 --- a/problems/0654.最大二叉树.md +++ b/problems/0654.最大二叉树.md @@ -292,6 +292,8 @@ class Solution: ``` (版本二) 使用下标 +```python + class Solution: def traversal(self, nums: List[int], left: int, right: int) -> TreeNode: if left >= right: @@ -309,7 +311,8 @@ class Solution: return self.traversal(nums, 0, len(nums)) -``` +```python + (版本三) 使用切片 class Solution: def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode: From 41cf3a4cdd276f52e2039a886ed21f0bdfa50444 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 22 May 2023 21:49:22 -0500 Subject: [PATCH 141/154] =?UTF-8?q?Update=200654.=E6=9C=80=E5=A4=A7?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0654.最大二叉树.md | 5 +++-- 1 file changed, 3 insertions(+), 2 deletions(-) diff --git a/problems/0654.最大二叉树.md b/problems/0654.最大二叉树.md index 07fbeb82..980523f5 100644 --- a/problems/0654.最大二叉树.md +++ b/problems/0654.最大二叉树.md @@ -292,7 +292,6 @@ class Solution: ``` (版本二) 使用下标 -```python class Solution: def traversal(self, nums: List[int], left: int, right: int) -> TreeNode: @@ -311,9 +310,11 @@ class Solution: return self.traversal(nums, 0, len(nums)) -```python (版本三) 使用切片 + +```python + class Solution: def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode: if not nums: From 7a544d95b8aa54cbb3e7457458fd4a67304048ae Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Mon, 22 May 2023 21:50:17 -0500 Subject: [PATCH 142/154] =?UTF-8?q?Update=200654.=E6=9C=80=E5=A4=A7?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0654.最大二叉树.md | 3 ++- 1 file changed, 2 insertions(+), 1 deletion(-) diff --git a/problems/0654.最大二叉树.md b/problems/0654.最大二叉树.md index 980523f5..33a9176e 100644 --- a/problems/0654.最大二叉树.md +++ b/problems/0654.最大二叉树.md @@ -292,6 +292,7 @@ class Solution: ``` (版本二) 使用下标 +```python class Solution: def traversal(self, nums: List[int], left: int, right: int) -> TreeNode: @@ -309,7 +310,7 @@ class Solution: def constructMaximumBinaryTree(self, nums: List[int]) -> TreeNode: return self.traversal(nums, 0, len(nums)) - + ``` (版本三) 使用切片 From 3058654e93415b39469468dabf4ac02907d04b71 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 20:47:45 -0500 Subject: [PATCH 143/154] =?UTF-8?q?Update=200098.=E9=AA=8C=E8=AF=81?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0098.验证二叉搜索树.md | 182 ++++++++++++------------- 1 file changed, 89 insertions(+), 93 deletions(-) diff --git a/problems/0098.验证二叉搜索树.md b/problems/0098.验证二叉搜索树.md index 95afe680..f2e148c2 100644 --- a/problems/0098.验证二叉搜索树.md +++ b/problems/0098.验证二叉搜索树.md @@ -341,117 +341,113 @@ class Solution { ## Python -**递归** - 利用BST中序遍历特性,把树"压缩"成数组 +递归法（版本一）利用中序递增性质，转换成数组 ```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: - def isValidBST(self, root: TreeNode) -> bool: - # 思路: 利用BST中序遍历的特性. - # 中序遍历输出的二叉搜索树节点的数值是有序序列 - candidate_list = [] - - def __traverse(root: TreeNode) -> None: - nonlocal candidate_list - if not root: - return - __traverse(root.left) - candidate_list.append(root.val) - __traverse(root.right) - - def __is_sorted(nums: list) -> bool: - for i in range(1, len(nums)): - if nums[i] <= nums[i - 1]: # ⚠️ 注意: Leetcode定义二叉搜索树中不能有重复元素 - return False - return True - - __traverse(root) - res = __is_sorted(candidate_list) - - return res -``` + def __init__(self): + self.vec = [] -**递归** - 标准做法 + def traversal(self, root): + if root is None: + return + self.traversal(root.left) + self.vec.append(root.val) # 将二叉搜索树转换为有序数组 + self.traversal(root.right) -```python -class Solution: - def isValidBST(self, root: TreeNode) -> bool: - # 规律: BST的中序遍历节点数值是从小到大. - cur_max = -float("INF") - def __isValidBST(root: TreeNode) -> bool: - nonlocal cur_max - - if not root: - return True - - is_left_valid = __isValidBST(root.left) - if cur_max < root.val: - cur_max = root.val - else: + def isValidBST(self, root): + self.vec = [] # 清空数组 + self.traversal(root) + for i in range(1, len(self.vec)): + # 注意要小于等于，搜索树里不能有相同元素 + if self.vec[i] <= self.vec[i - 1]: return False - is_right_valid = __isValidBST(root.right) - - return is_left_valid and is_right_valid - return __isValidBST(root) + return True + ``` -**递归** - 避免初始化最小值做法： + +递归法（版本二）设定极小值，进行比较 + ```python class Solution: - def isValidBST(self, root: TreeNode) -> bool: - # 规律: BST的中序遍历节点数值是从小到大. - pre = None - def __isValidBST(root: TreeNode) -> bool: - nonlocal pre - - if not root: - return True - - is_left_valid = __isValidBST(root.left) - if pre and pre.val>=root.val: return False - pre = root - is_right_valid = __isValidBST(root.right) - - return is_left_valid and is_right_valid - return __isValidBST(root) + def __init__(self): + self.maxVal = float('-inf') # 因为后台测试数据中有int最小值 + + def isValidBST(self, root): + if root is None: + return True + + left = self.isValidBST(root.left) + # 中序遍历，验证遍历的元素是不是从小到大 + if self.maxVal < root.val: + self.maxVal = root.val + else: + return False + right = self.isValidBST(root.right) + + return left and right + ``` +递归法（版本三）直接取该树的最小值 ```python -迭代-中序遍历 +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right class Solution: - def isValidBST(self, root: TreeNode) -> bool: + def __init__(self): + self.pre = None # 用来记录前一个节点 + + def isValidBST(self, root): + if root is None: + return True + + left = self.isValidBST(root.left) + + if self.pre is not None and self.pre.val >= root.val: + return False + self.pre = root # 记录前一个节点 + + right = self.isValidBST(root.right) + return left and right + + + +``` +迭代法 +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def isValidBST(self, root): stack = [] cur = root - pre = None - while cur or stack: - if cur: # 指针来访问节点，访问到最底层 + pre = None # 记录前一个节点 + while cur is not None or len(stack) > 0: + if cur is not None: stack.append(cur) - cur = cur.left - else: # 逐一处理节点 - cur = stack.pop() - if pre and cur.val <= pre.val: # 比较当前节点和前节点的值的大小 + cur = cur.left # 左 + else: + cur = stack.pop() # 中 + if pre is not None and cur.val <= pre.val: return False - pre = cur - cur = cur.right + pre = cur # 保存前一个访问的结点 + cur = cur.right # 右 return True ``` -```python -# 遵循Carl的写法，只添加了节点判断的部分 -class Solution: - def isValidBST(self, root: TreeNode) -> bool: - # method 2 - que, pre = [], None - while root or que: - while root: - que.append(root) - root = root.left - root = que.pop() - # 对第一个节点只做记录，对后面的节点进行比较 - if pre is None: - pre = root.val - else: - if pre >= root.val: return False - pre = root.val - root = root.right - return True -``` + ## Go From 85e4c414571db69f8dd4c27c67928632845e65e4 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 20:56:07 -0500 Subject: [PATCH 144/154] =?UTF-8?q?Update=200530.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=90=9C=E7=B4=A2=E6=A0=91=E7=9A=84=E6=9C=80=E5=B0=8F=E7=BB=9D?= =?UTF-8?q?=E5=AF=B9=E5=B7=AE.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0530.二叉搜索树的最小绝对差.md | 100 ++++++++++-------- 1 file changed, 58 insertions(+), 42 deletions(-) diff --git a/problems/0530.二叉搜索树的最小绝对差.md b/problems/0530.二叉搜索树的最小绝对差.md index fa1430de..3fe57702 100644 --- a/problems/0530.二叉搜索树的最小绝对差.md +++ b/problems/0530.二叉搜索树的最小绝对差.md @@ -204,66 +204,82 @@ class Solution { ``` ## Python -递归 +递归法（版本一）利用中序递增，结合数组 ```python class Solution: - def getMinimumDifference(self, root: TreeNode) -> int: - res = [] - r = float("inf") - def buildaList(root): //把二叉搜索树转换成有序数组 - if not root: return None - if root.left: buildaList(root.left) //左 - res.append(root.val) //中 - if root.right: buildaList(root.right) //右 - return res - - buildaList(root) - for i in range(len(res)-1): // 统计有序数组的最小差值 - r = min(abs(res[i]-res[i+1]),r) - return r - - -class Solution: # 双指针法，不用数组 (同Carl写法) - 更快 - def getMinimumDifference(self, root: Optional[TreeNode]) -> int: - global pre,minval - pre = None - minval = 10**5 - self.traversal(root) - return minval + def __init__(self): + self.vec = [] - def traversal(self,root): - global pre,minval - if not root: return None + def traversal(self, root): + if root is None: + return self.traversal(root.left) - if pre and root.val-pre.val int: + def __init__(self): + self.result = float('inf') + self.pre = None + + def traversal(self, cur): + if cur is None: + return + self.traversal(cur.left) # 左 + if self.pre is not None: # 中 + self.result = min(self.result, cur.val - self.pre.val) + self.pre = cur # 记录前一个 + self.traversal(cur.right) # 右 + + def getMinimumDifference(self, root): + self.traversal(root) + return self.result + + +``` + +迭代法 +```python +class Solution: + def getMinimumDifference(self, root): stack = [] cur = root pre = None result = float('inf') - while cur or stack: - if cur: # 指针来访问节点，访问到最底层 - stack.append(cur) - cur = cur.left - else: # 逐一处理节点 + + while cur is not None or len(stack) > 0: + if cur is not None: + stack.append(cur) # 将访问的节点放进栈 + cur = cur.left # 左 + else: cur = stack.pop() - if pre: # 当前节点和前节点的值的差值 - result = min(result, abs(cur.val - pre.val)) + if pre is not None: # 中 + result = min(result, cur.val - pre.val) pre = cur - cur = cur.right + cur = cur.right # 右 + return result + + ``` - ## Go： 中序遍历，然后计算最小差值 From c5d1845f3f48ff7ccd78319fdb4481033b159df1 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 21:06:32 -0500 Subject: [PATCH 145/154] =?UTF-8?q?Update=200501.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84=E4=BC=97=E6=95=B0?= =?UTF-8?q?.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0501.二叉搜索树中的众数.md | 144 +++++++++++-------- 1 file changed, 87 insertions(+), 57 deletions(-) diff --git a/problems/0501.二叉搜索树中的众数.md b/problems/0501.二叉搜索树中的众数.md index b7ef606f..6dc8ed83 100644 --- a/problems/0501.二叉搜索树中的众数.md +++ b/problems/0501.二叉搜索树中的众数.md @@ -475,8 +475,7 @@ class Solution { ## Python -> 递归法 -> 常量空间，递归产生的栈不算 +递归法（版本一）利用字典 ```python # Definition for a binary tree node. @@ -485,77 +484,108 @@ class Solution { # self.val = val # self.left = left # self.right = right +from collections import defaultdict + class Solution: - def __init__(self): - self.pre = TreeNode() - self.count = 0 - self.max_count = 0 - self.result = [] + def searchBST(self, cur, freq_map): + if cur is None: + return + freq_map[cur.val] += 1 # 统计元素频率 + self.searchBST(cur.left, freq_map) + self.searchBST(cur.right, freq_map) - def findMode(self, root: TreeNode) -> List[int]: - if not root: return None - self.search_BST(root) - return self.result - - def search_BST(self, cur: TreeNode) -> None: - if not cur: return None - self.search_BST(cur.left) - # 第一个节点 - if not self.pre: - self.count = 1 - # 与前一个节点数值相同 - elif self.pre.val == cur.val: - self.count += 1 - # 与前一个节点数值不相同 - else: - self.count = 1 - self.pre = cur + def findMode(self, root): + freq_map = defaultdict(int) # key:元素，value:出现频率 + result = [] + if root is None: + return result + self.searchBST(root, freq_map) + max_freq = max(freq_map.values()) + for key, freq in freq_map.items(): + if freq == max_freq: + result.append(key) + return result - if self.count == self.max_count: - self.result.append(cur.val) - - if self.count > self.max_count: - self.max_count = self.count - self.result = [cur.val] # 清空self.result，确保result之前的的元素都失效 - - self.search_BST(cur.right) ``` +递归法（版本二）利用二叉搜索树性质 -> 迭代法-中序遍历 -> 利用二叉搜索树特性，在历遍过程中更新结果，一次历遍 -> 但需要使用额外空间存储历遍的节点 ```python class Solution: - def findMode(self, root: TreeNode) -> List[int]: - stack = [] + def __init__(self): + self.maxCount = 0 # 最大频率 + self.count = 0 # 统计频率 + self.pre = None + self.result = [] + + def searchBST(self, cur): + if cur is None: + return + + self.searchBST(cur.left) # 左 + # 中 + if self.pre is None: # 第一个节点 + self.count = 1 + elif self.pre.val == cur.val: # 与前一个节点数值相同 + self.count += 1 + else: # 与前一个节点数值不同 + self.count = 1 + self.pre = cur # 更新上一个节点 + + if self.count == self.maxCount: # 如果与最大值频率相同，放进result中 + self.result.append(cur.val) + + if self.count > self.maxCount: # 如果计数大于最大值频率 + self.maxCount = self.count # 更新最大频率 + self.result = [cur.val] # 很关键的一步，不要忘记清空result，之前result里的元素都失效了 + + self.searchBST(cur.right) # 右 + return + + def findMode(self, root): + self.count = 0 + self.maxCount = 0 + self.pre = None # 记录前一个节点 + self.result = [] + + self.searchBST(root) + return self.result +``` +迭代法 +```python +class Solution: + def findMode(self, root): + st = [] cur = root pre = None - maxCount, count = 0, 0 - res = [] - while cur or stack: - if cur: # 指针来访问节点，访问到最底层 - stack.append(cur) - cur = cur.left - else: # 逐一处理节点 - cur = stack.pop() - if pre == None: # 第一个节点 + maxCount = 0 # 最大频率 + count = 0 # 统计频率 + result = [] + + while cur is not None or st: + if cur is not None: # 指针来访问节点，访问到最底层 + st.append(cur) # 将访问的节点放进栈 + cur = cur.left # 左 + else: + cur = st.pop() + if pre is None: # 第一个节点 count = 1 elif pre.val == cur.val: # 与前一个节点数值相同 count += 1 - else: + else: # 与前一个节点数值不同 count = 1 - if count == maxCount: - res.append(cur.val) - if count > maxCount: - maxCount = count - res.clear() - res.append(cur.val) + + if count == maxCount: # 如果和最大值相同，放进result中 + result.append(cur.val) + + if count > maxCount: # 如果计数大于最大值频率 + maxCount = count # 更新最大频率 + result = [cur.val] # 很关键的一步，不要忘记清空result，之前result里的元素都失效了 pre = cur - cur = cur.right - return res - + cur = cur.right # 右 + + return result ``` ## Go From daa5417db9ef711d219ca8e201c6dd0cd57c9a20 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 21:11:37 -0500 Subject: [PATCH 146/154] =?UTF-8?q?Update=200236.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E8=BF=91=E5=85=AC=E5=85=B1=E7=A5=96?= =?UTF-8?q?=E5=85=88.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0236.二叉树的最近公共祖先.md | 45 +++++++++++++------ 1 file changed, 32 insertions(+), 13 deletions(-) diff --git a/problems/0236.二叉树的最近公共祖先.md b/problems/0236.二叉树的最近公共祖先.md index 33201def..0ebd5566 100644 --- a/problems/0236.二叉树的最近公共祖先.md +++ b/problems/0236.二叉树的最近公共祖先.md @@ -274,25 +274,44 @@ class Solution { ``` ## Python - +递归法（版本一） ```python class Solution: - """二叉树的最近公共祖先 递归法""" - - def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': - if not root or root == p or root == q: + def lowestCommonAncestor(self, root, p, q): + if root == q or root == p or root is None: return root - + left = self.lowestCommonAncestor(root.left, p, q) right = self.lowestCommonAncestor(root.right, p, q) - - if left and right: - return root - if left: - return left - return right -``` + if left is not None and right is not None: + return root + + if left is None and right is not None: + return right + elif left is not None and right is None: + return left + else: + return None +``` +递归法（版本二）精简 +```python +class Solution: + def lowestCommonAncestor(self, root, p, q): + if root == q or root == p or root is None: + return root + + left = self.lowestCommonAncestor(root.left, p, q) + right = self.lowestCommonAncestor(root.right, p, q) + + if left is not None and right is not None: + return root + + if left is None: + return right + return left + +``` ## Go ```Go From e7b7aa8e4a85d49933ddfa14e4911cc8d0f72bfe Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 21:16:37 -0500 Subject: [PATCH 147/154] =?UTF-8?q?Update=200235.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=90=9C=E7=B4=A2=E6=A0=91=E7=9A=84=E6=9C=80=E8=BF=91=E5=85=AC?= =?UTF-8?q?=E5=85=B1=E7=A5=96=E5=85=88.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...35.二叉搜索树的最近公共祖先.md | 49 ++++++++++++++----- 1 file changed, 36 insertions(+), 13 deletions(-) diff --git a/problems/0235.二叉搜索树的最近公共祖先.md b/problems/0235.二叉搜索树的最近公共祖先.md index 8353303a..9777bb0b 100644 --- a/problems/0235.二叉搜索树的最近公共祖先.md +++ b/problems/0235.二叉搜索树的最近公共祖先.md @@ -275,34 +275,57 @@ class Solution { ## Python -递归法： +递归法（版本一） ```python class Solution: - """二叉搜索树的最近公共祖先 递归法""" + def traversal(self, cur, p, q): + if cur is None: + return cur + # 中 + if cur.val > p.val and cur.val > q.val: # 左 + left = self.traversal(cur.left, p, q) + if left is not None: + return left - def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': - if root.val > p.val and root.val > q.val: - return self.lowestCommonAncestor(root.left, p, q) - if root.val < p.val and root.val < q.val: - return self.lowestCommonAncestor(root.right, p, q) - return root + if cur.val < p.val and cur.val < q.val: # 右 + right = self.traversal(cur.right, p, q) + if right is not None: + return right + + return cur + + def lowestCommonAncestor(self, root, p, q): + return self.traversal(root, p, q) ``` -迭代法： +迭代法（版本二）精简 ```python class Solution: - """二叉搜索树的最近公共祖先 迭代法""" + def lowestCommonAncestor(self, root, p, q): + if root.val > p.val and root.val > q.val: + return self.lowestCommonAncestor(root.left, p, q) + elif root.val < p.val and root.val < q.val: + return self.lowestCommonAncestor(root.right, p, q) + else: + return root - def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode': - while True: +``` + +迭代法 +```python +class Solution: + def lowestCommonAncestor(self, root, p, q): + while root: if root.val > p.val and root.val > q.val: root = root.left elif root.val < p.val and root.val < q.val: root = root.right else: return root -``` + return None + +``` ## Go 递归法： From 511bf447936dbb628609c5b1cad23462f20ac817 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 21:29:16 -0500 Subject: [PATCH 148/154] =?UTF-8?q?Update=200701.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84=E6=8F=92=E5=85=A5?= =?UTF-8?q?=E6=93=8D=E4=BD=9C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0701.二叉搜索树中的插入操作.md | 189 ++++++++---------- 1 file changed, 80 insertions(+), 109 deletions(-) diff --git a/problems/0701.二叉搜索树中的插入操作.md b/problems/0701.二叉搜索树中的插入操作.md index 4e834201..1ba7461f 100644 --- a/problems/0701.二叉搜索树中的插入操作.md +++ b/problems/0701.二叉搜索树中的插入操作.md @@ -256,132 +256,103 @@ class Solution { ----- ## Python -**递归法** - 有返回值 - +递归法（版本一） ```python -# Definition for a binary tree node. -# class TreeNode: -# def __init__(self, val=0, left=None, right=None): -# self.val = val -# self.left = left -# self.right = right class Solution: - def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode: - # 返回更新后的以当前root为根节点的新树，方便用于更新上一层的父子节点关系链 + def __init__(self): + self.parent = None - # Base Case - if not root: return TreeNode(val) + def traversal(self, cur, val): + if cur is None: + node = TreeNode(val) + if val > self.parent.val: + self.parent.right = node + else: + self.parent.left = node + return - # 单层递归逻辑: - if val < root.val: - # 将val插入至当前root的左子树中合适的位置 - # 并更新当前root的左子树为包含目标val的新左子树 - root.left = self.insertIntoBST(root.left, val) + self.parent = cur + if cur.val > val: + self.traversal(cur.left, val) + if cur.val < val: + self.traversal(cur.right, val) - if root.val < val: - # 将val插入至当前root的右子树中合适的位置 - # 并更新当前root的右子树为包含目标val的新右子树 - root.right = self.insertIntoBST(root.right, val) - - # 返回更新后的以当前root为根节点的新树 + def insertIntoBST(self, root, val): + self.parent = TreeNode(0) + if root is None: + return TreeNode(val) + self.traversal(root, val) return root + ``` -**递归法** - 无返回值 +递归法（版本二） ```python class Solution: - def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode: - if not root: + def insertIntoBST(self, root, val): + if root is None: return TreeNode(val) parent = None - def __traverse(cur: TreeNode, val: int) -> None: - # 在函数运行的同时把新节点插入到该被插入的地方. - nonlocal parent - if not cur: - new_node = TreeNode(val) - if parent.val < val: - parent.right = new_node - else: - parent.left = new_node - return - - parent = cur # 重点: parent的作用只有运行到上面if not cur:才会发挥出来. - if cur.val < val: - __traverse(cur.right, val) - else: - __traverse(cur.left, val) - return - __traverse(root, val) - return root -``` - -**递归法** - 无返回值 - another easier way -```python -class Solution: - def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]: - newNode = TreeNode(val) - if not root: return newNode - - if not root.left and val < root.val: - root.left = newNode - if not root.right and val > root.val: - root.right = newNode - - if val < root.val: - self.insertIntoBST(root.left, val) - if val > root.val: - self.insertIntoBST(root.right, val) - - return root -``` - -**递归法** - 无返回值 有注释 不用Helper function -```python -class Solution: - def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]: - if not root: # for root==None - return TreeNode(val) - if root.valval: - if root.left==None: # found the parent - root.left = TreeNode(val) - else: # not found, keep searching - self.insertIntoBST(root.left, val) - # return the final tree - return root -``` - -**迭代法** -与无返回值的递归函数的思路大体一致 -```python -class Solution: - def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode: - if not root: - return TreeNode(val) - parent = None # 此步可以省略 cur = root - - # 用while循环不断地找新节点的parent - while cur: - parent = cur # 首先保存当前非空节点作为下一次迭代的父节点 - if cur.val < val: - cur = cur.right - elif cur.val > val: + while cur: + parent = cur + if val < cur.val: cur = cur.left - - # 运行到这意味着已经跳出上面的while循环, - # 同时意味着新节点的parent已经被找到. - # parent已被找到, 新节点已经ready. 把两个节点黏在一起就好了. - if parent.val > val: + else: + cur = cur.right + if val < parent.val: parent.left = TreeNode(val) - else: + else: parent.right = TreeNode(val) - return root +``` + +递归法（版本三） +```python +class Solution: + def insertIntoBST(self, root: Optional[TreeNode], val: int) -> Optional[TreeNode]: + if root is None or root.val == val: + return TreeNode(val) + elif root.val > val: + if root.left is None: + root.left = TreeNode(val) + else: + self.insertIntoBST(root.left, val) + elif root.val < val: + if root.right is None: + root.right = TreeNode(val) + else: + self.insertIntoBST(root.right, val) + return root +``` + + + +迭代法 +```python +class Solution: + def insertIntoBST(self, root, val): + if root is None: # 如果根节点为空，创建新节点作为根节点并返回 + node = TreeNode(val) + return node + + cur = root + parent = root # 记录上一个节点，用于连接新节点 + while cur is not None: + parent = cur + if cur.val > val: + cur = cur.left + else: + cur = cur.right + + node = TreeNode(val) + if val < parent.val: + parent.left = node # 将新节点连接到父节点的左子树 + else: + parent.right = node # 将新节点连接到父节点的右子树 + + return root + ``` ----- From 731d1ca8aa61b1b8969ead0c5cf8b7d8b21b55e6 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 21:38:27 -0500 Subject: [PATCH 149/154] =?UTF-8?q?Update=200450.=E5=88=A0=E9=99=A4?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84?= =?UTF-8?q?=E8=8A=82=E7=82=B9.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0450.删除二叉搜索树中的节点.md | 132 +++++++++--------- 1 file changed, 67 insertions(+), 65 deletions(-) diff --git a/problems/0450.删除二叉搜索树中的节点.md b/problems/0450.删除二叉搜索树中的节点.md index 2602b528..d4dca1d4 100644 --- a/problems/0450.删除二叉搜索树中的节点.md +++ b/problems/0450.删除二叉搜索树中的节点.md @@ -324,88 +324,90 @@ class Solution { ``` ## Python - +递归法（版本一） ```python class Solution: - def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]: - if not root : return None # 节点为空，返回 - if root.val < key : - root.right = self.deleteNode(root.right, key) - elif root.val > key : + def deleteNode(self, root, key): + if root is None: + return root + if root.val == key: + if root.left is None and root.right is None: + return None + elif root.left is None: + return root.right + elif root.right is None: + return root.left + else: + cur = root.right + while cur.left is not None: + cur = cur.left + cur.left = root.left + return root.right + if root.val > key: root.left = self.deleteNode(root.left, key) - else: - # 当前节点的左子树为空，返回当前的右子树 - if not root.left : return root.right - # 当前节点的右子树为空，返回当前的左子树 - if not root.right: return root.left - # 左右子树都不为空，找到右孩子的最左节点 记为p - node = root.right - while node.left : - node = node.left - # 将当前节点的左子树挂在p的左孩子上 - node.left = root.left - # 当前节点的右子树替换掉当前节点，完成当前节点的删除 - root = root.right + if root.val < key: + root.right = self.deleteNode(root.right, key) return root ``` -**普通二叉树的删除方式** +递归法（版本二） ```python class Solution: - def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]: - if not root: return root - if root.val == key: - if not root.right: # 这里第二次操作目标值：最终删除的作用 + def deleteNode(self, root, key): + if root is None: # 如果根节点为空，直接返回 + return root + if root.val == key: # 找到要删除的节点 + if root.right is None: # 如果右子树为空，直接返回左子树作为新的根节点 return root.left - tmp = root.right - while tmp.left: - tmp = tmp.left - root.val, tmp.val = tmp.val, root.val # 这里第一次操作目标值：交换目标值其右子树最左面节点。 - - root.left = self.deleteNode(root.left, key) - root.right = self.deleteNode(root.right, key) + cur = root.right + while cur.left: # 找到右子树中的最左节点 + cur = cur.left + root.val, cur.val = cur.val, root.val # 将要删除的节点值与最左节点值交换 + root.left = self.deleteNode(root.left, key) # 在左子树中递归删除目标节点 + root.right = self.deleteNode(root.right, key) # 在右子树中递归删除目标节点 return root + ``` **迭代法** ```python class Solution: - def deleteNode(self, root: Optional[TreeNode], key: int) -> Optional[TreeNode]: - # 找到节点后分两步，1. 把节点的左子树和右子树连起来，2. 把右子树跟父节点连起来 - # root is None - if not root: return root - p = root - last = None - while p: - if p.val==key: - # 1. connect left to right - # right is not None -> left is None | left is not None - if p.right: - if p.left: - node = p.right - while node.left: - node = node.left - node.left = p.left - right = p.right - else: - # right is None -> right=left - right = p.left - # 2. connect right to last - if last==None: - root = right - elif last.val>key: - last.left = right - else: - last.right = right - # 3. return + def deleteOneNode(self, target: TreeNode) -> TreeNode: + """ + 将目标节点（删除节点）的左子树放到目标节点的右子树的最左面节点的左孩子位置上 + 并返回目标节点右孩子为新的根节点 + 是动画里模拟的过程 + """ + if target is None: + return target + if target.right is None: + return target.left + cur = target.right + while cur.left: + cur = cur.left + cur.left = target.left + return target.right + + def deleteNode(self, root: TreeNode, key: int) -> TreeNode: + if root is None: + return root + cur = root + pre = None # 记录cur的父节点，用来删除cur + while cur: + if cur.val == key: break + pre = cur + if cur.val > key: + cur = cur.left else: - # Update last and continue - last = p - if p.val>key: - p = p.left - else: - p = p.right + cur = cur.right + if pre is None: # 如果搜索树只有头结点 + return self.deleteOneNode(cur) + # pre 要知道是删左孩子还是右孩子 + if pre.left and pre.left.val == key: + pre.left = self.deleteOneNode(cur) + if pre.right and pre.right.val == key: + pre.right = self.deleteOneNode(cur) return root ``` From 53652744d2aa05765a10ce824424ee1c06afc57c Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 21:43:40 -0500 Subject: [PATCH 150/154] =?UTF-8?q?Update=200669.=E4=BF=AE=E5=89=AA?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0669.修剪二叉搜索树.md | 86 ++++++++++++++------------ 1 file changed, 47 insertions(+), 39 deletions(-) diff --git a/problems/0669.修剪二叉搜索树.md b/problems/0669.修剪二叉搜索树.md index 18d8a0cc..95372d61 100644 --- a/problems/0669.修剪二叉搜索树.md +++ b/problems/0669.修剪二叉搜索树.md @@ -271,64 +271,72 @@ class Solution { ## Python -**递归** - +递归法（版本一） ```python -# Definition for a binary tree node. -# class TreeNode: -# def __init__(self, val=0, left=None, right=None): -# self.val = val -# self.left = left -# self.right = right class Solution: def trimBST(self, root: TreeNode, low: int, high: int) -> TreeNode: - ''' - 确认递归函数参数以及返回值：返回更新后剪枝后的当前root节点 - ''' - # Base Case - if not root: return None - - # 单层递归逻辑 + if root is None: + return None if root.val < low: - # 若当前root节点小于左界：只考虑其右子树，用于替代更新后的其本身，抛弃其左子树整体 + # 寻找符合区间 [low, high] 的节点 return self.trimBST(root.right, low, high) - - if high < root.val: - # 若当前root节点大于右界：只考虑其左子树，用于替代更新后的其本身，抛弃其右子树整体 + if root.val > high: + # 寻找符合区间 [low, high] 的节点 return self.trimBST(root.left, low, high) + root.left = self.trimBST(root.left, low, high) # root.left 接入符合条件的左孩子 + root.right = self.trimBST(root.right, low, high) # root.right 接入符合条件的右孩子 + return root - if low <= root.val <= high: - root.left = self.trimBST(root.left, low, high) - root.right = self.trimBST(root.right, low, high) - # 返回更新后的剪枝过的当前节点root - return root ``` - -**迭代** - +递归法（版本二）精简 ```python class Solution: - def trimBST(self, root: Optional[TreeNode], low: int, high: int) -> Optional[TreeNode]: - if not root: return root - # 处理头结点，让root移动到[L, R] 范围内，注意是左闭右开 - while root and (root.val < low or root.val > high): - if root.val < low: # 小于L往右走 - root = root.right - else: # 大于R往左走 - root = root.left - # 此时root已经在[L, R] 范围内，处理左孩子元素小于L的情况 + def trimBST(self, root: TreeNode, low: int, high: int) -> TreeNode: + if root is None: + return None + if root.val < low: + return self.trimBST(root.right, low, high) + if root.val > high: + return self.trimBST(root.left, low, high) + root.left = self.trimBST(root.left, low, high) + root.right = self.trimBST(root.right, low, high) + return root + + +``` + +迭代法 +```python +class Solution: + def trimBST(self, root: TreeNode, L: int, R: int) -> TreeNode: + if not root: + return None + + # 处理头结点，让root移动到[L, R] 范围内，注意是左闭右闭 + while root and (root.val < L or root.val > R): + if root.val < L: + root = root.right # 小于L往右走 + else: + root = root.left # 大于R往左走 + cur = root + + # 此时root已经在[L, R] 范围内，处理左孩子元素小于L的情况 while cur: - while cur.left and cur.left.val < low: + while cur.left and cur.left.val < L: cur.left = cur.left.right cur = cur.left - # 此时root已经在[L, R] 范围内，处理右孩子大于R的情况 + cur = root + + # 此时root已经在[L, R] 范围内，处理右孩子大于R的情况 while cur: - while cur.right and cur.right.val > high: + while cur.right and cur.right.val > R: cur.right = cur.right.left cur = cur.right + return root + ``` ## Go From 76e3811c5e7d2299ff06c2aeb231f8fa78bcb957 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 21:46:41 -0500 Subject: [PATCH 151/154] =?UTF-8?q?Update=200108.=E5=B0=86=E6=9C=89?= =?UTF-8?q?=E5=BA=8F=E6=95=B0=E7=BB=84=E8=BD=AC=E6=8D=A2=E4=B8=BA=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...将有序数组转换为二叉搜索树.md | 98 +++++++++---------- 1 file changed, 45 insertions(+), 53 deletions(-) diff --git a/problems/0108.将有序数组转换为二叉搜索树.md b/problems/0108.将有序数组转换为二叉搜索树.md index 89778421..056ef3e2 100644 --- a/problems/0108.将有序数组转换为二叉搜索树.md +++ b/problems/0108.将有序数组转换为二叉搜索树.md @@ -316,73 +316,65 @@ class Solution { ``` ## Python -**递归** - +递归法 ```python -# Definition for a binary tree node. -# class TreeNode: -# def __init__(self, val=0, left=None, right=None): -# self.val = val -# self.left = left -# self.right = right class Solution: - def sortedArrayToBST(self, nums: List[int]) -> TreeNode: - ''' - 构造二叉树：重点是选取数组最中间元素为分割点，左侧是递归左区间；右侧是递归右区间 - 必然是平衡树 - 左闭右闭区间 - ''' - # 返回根节点 - root = self.traversal(nums, 0, len(nums)-1) - return root - def traversal(self, nums: List[int], left: int, right: int) -> TreeNode: - # Base Case if left > right: return None - # 确定左右界的中心，防越界 mid = left + (right - left) // 2 - # 构建根节点 - mid_root = TreeNode(nums[mid]) - # 构建以左右界的中心为分割点的左右子树 - mid_root.left = self.traversal(nums, left, mid-1) - mid_root.right = self.traversal(nums, mid+1, right) + root = TreeNode(nums[mid]) + root.left = self.traversal(nums, left, mid - 1) + root.right = self.traversal(nums, mid + 1, right) + return root + + def sortedArrayToBST(self, nums: List[int]) -> TreeNode: + root = self.traversal(nums, 0, len(nums) - 1) + return root - # 返回由被传入的左右界定义的某子树的根节点 - return mid_root ``` -**迭代**（左闭右开） +迭代法 ```python +from collections import deque + class Solution: - def sortedArrayToBST(self, nums: List[int]) -> Optional[TreeNode]: - if len(nums) == 0: return None - root = TreeNode() # 初始化 - nodeSt = [root] - leftSt = [0] - rightSt = [len(nums)] - - while nodeSt: - node = nodeSt.pop() # 处理根节点 - left = leftSt.pop() - right = rightSt.pop() - mid = left + (right - left) // 2 - node.val = nums[mid] - - if left < mid: # 处理左区间 - node.left = TreeNode() - nodeSt.append(node.left) - leftSt.append(left) - rightSt.append(mid) - - if right > mid + 1: # 处理右区间 - node.right = TreeNode() - nodeSt.append(node.right) - leftSt.append(mid + 1) - rightSt.append(right) + def sortedArrayToBST(self, nums: List[int]) -> TreeNode: + if len(nums) == 0: + return None + root = TreeNode(0) # 初始根节点 + nodeQue = deque() # 放遍历的节点 + leftQue = deque() # 保存左区间下标 + rightQue = deque() # 保存右区间下标 + + nodeQue.append(root) # 根节点入队列 + leftQue.append(0) # 0为左区间下标初始位置 + rightQue.append(len(nums) - 1) # len(nums) - 1为右区间下标初始位置 + + while nodeQue: + curNode = nodeQue.popleft() + left = leftQue.popleft() + right = rightQue.popleft() + mid = left + (right - left) // 2 + + curNode.val = nums[mid] # 将mid对应的元素给中间节点 + + if left <= mid - 1: # 处理左区间 + curNode.left = TreeNode(0) + nodeQue.append(curNode.left) + leftQue.append(left) + rightQue.append(mid - 1) + + if right >= mid + 1: # 处理右区间 + curNode.right = TreeNode(0) + nodeQue.append(curNode.right) + leftQue.append(mid + 1) + rightQue.append(right) + return root + ``` ## Go From 14acf5a55469e766e05b74f2279132ce2a7b66d7 Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Tue, 23 May 2023 21:57:22 -0500 Subject: [PATCH 152/154] =?UTF-8?q?Update=200538.=E6=8A=8A=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E8=BD=AC=E6=8D=A2=E4=B8=BA?= =?UTF-8?q?=E7=B4=AF=E5=8A=A0=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...38.把二叉搜索树转换为累加树.md | 53 +++++++++++++++++-- 1 file changed, 50 insertions(+), 3 deletions(-) diff --git a/problems/0538.把二叉搜索树转换为累加树.md b/problems/0538.把二叉搜索树转换为累加树.md index ad5310e1..ad4decc5 100644 --- a/problems/0538.把二叉搜索树转换为累加树.md +++ b/problems/0538.把二叉搜索树转换为累加树.md @@ -200,8 +200,30 @@ class Solution { ``` ## Python -**递归** - +递归法(版本一) +```python +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def convertBST(self, root: TreeNode) -> TreeNode: + self.pre = 0 # 记录前一个节点的数值 + self.traversal(root) + return root + def traversal(self, cur): + if cur is None: + return + self.traversal(cur.right) + cur.val += self.pre + self.pre = cur.val + self.traversal(cur.left) + + +``` +递归法（版本二） ```python # Definition for a binary tree node. # class TreeNode: @@ -234,7 +256,32 @@ class Solution: return root ``` -**迭代** +迭代法（版本一） +```python +class Solution: + def __init__(self): + self.pre = 0 # 记录前一个节点的数值 + + def traversal(self, root): + stack = [] + cur = root + while cur or stack: + if cur: + stack.append(cur) + cur = cur.right # 右 + else: + cur = stack.pop() # 中 + cur.val += self.pre + self.pre = cur.val + cur = cur.left # 左 + + def convertBST(self, root): + self.pre = 0 + self.traversal(root) + return root + +``` +迭代法（版本二） ```python class Solution: def convertBST(self, root: Optional[TreeNode]) -> Optional[TreeNode]: From 450b3f2f15ab7afa7b18f835879dc1573249edee Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 24 May 2023 00:51:14 -0500 Subject: [PATCH 153/154] =?UTF-8?q?Update=200111.=E4=BA=8C=E5=8F=89?= =?UTF-8?q?=E6=A0=91=E7=9A=84=E6=9C=80=E5=B0=8F=E6=B7=B1=E5=BA=A6.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0111.二叉树的最小深度.md | 108 +++++++++++----------- 1 file changed, 56 insertions(+), 52 deletions(-) diff --git a/problems/0111.二叉树的最小深度.md b/problems/0111.二叉树的最小深度.md index 0c086f1b..ef59730e 100644 --- a/problems/0111.二叉树的最小深度.md +++ b/problems/0111.二叉树的最小深度.md @@ -297,36 +297,72 @@ class Solution { ## Python -递归法： +递归法（版本一） ```python -# Definition for a binary tree node. -# class TreeNode: -# def __init__(self, val=0, left=None, right=None): -# self.val = val -# self.left = left -# self.right = right class Solution: - def minDepth(self, root: TreeNode) -> int: - if not root: + def getDepth(self, node): + if node is None: return 0 + leftDepth = self.getDepth(node.left) # 左 + rightDepth = self.getDepth(node.right) # 右 - if not root.left and not root.right: - return 1 + # 当一个左子树为空，右不为空，这时并不是最低点 + if node.left is None and node.right is not None: + return 1 + rightDepth - left_depth = float('inf') - right_depth = float('inf') + # 当一个右子树为空，左不为空，这时并不是最低点 + if node.left is not None and node.right is None: + return 1 + leftDepth - if root.left: - left_depth = self.minDepth(root.left) - if root.right: - right_depth = self.minDepth(root.right) - - return 1 + min(left_depth, right_depth) + result = 1 + min(leftDepth, rightDepth) + return result + + def minDepth(self, root): + return self.getDepth(root) ``` +递归法（版本二） -迭代法： +```python +class Solution: + def minDepth(self, root): + if root is None: + return 0 + if root.left is None and root.right is not None: + return 1 + self.minDepth(root.right) + if root.left is not None and root.right is None: + return 1 + self.minDepth(root.left) + return 1 + min(self.minDepth(root.left), self.minDepth(root.right)) + + +``` +递归法（版本三）前序 + +```python +class Solution: + def __init__(self): + self.result = float('inf') + + def getDepth(self, node, depth): + if node is None: + return + if node.left is None and node.right is None: + self.result = min(self.result, depth) + if node.left: + self.getDepth(node.left, depth + 1) + if node.right: + self.getDepth(node.right, depth + 1) + + def minDepth(self, root): + if root is None: + return 0 + self.getDepth(root, 1) + return self.result + + +``` +迭代法 ```python # Definition for a binary tree node. @@ -359,39 +395,7 @@ class Solution: return depth ``` -迭代法： -```python -# Definition for a binary tree node. -# class TreeNode: -# def __init__(self, val=0, left=None, right=None): -# self.val = val -# self.left = left -# self.right = right - -class Solution: - def minDepth(self, root: TreeNode) -> int: - if not root: - return 0 - - queue = collections.deque([(root, 1)]) - - while queue: - node, depth = queue.popleft() - - # Check if the node is a leaf node - if not node.left and not node.right: - return depth - - # Add left and right child to the queue - if node.left: - queue.append((node.left, depth+1)) - if node.right: - queue.append((node.right, depth+1)) - - return 0 - -``` ## Go From 832897e4bbeb21c5906729257af347f6cd90c97c Mon Sep 17 00:00:00 2001 From: jianghongcheng <35664721+jianghongcheng@users.noreply.github.com> Date: Wed, 24 May 2023 01:56:01 -0500 Subject: [PATCH 154/154] =?UTF-8?q?Update=200110.=E5=B9=B3=E8=A1=A1?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 22 ++++++++++------------ 1 file changed, 10 insertions(+), 12 deletions(-) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index c0561e10..e10a612a 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -536,18 +536,16 @@ class Solution: ```python class Solution: - def isBalanced(self, root: TreeNode) -> bool: - return self.height(root) != -1 - def height(self, node: TreeNode) -> int: - if not node: - return 0 - left = self.height(node.left) - if left == -1: - return -1 - right = self.height(node.right) - if right == -1 or abs(left - right) > 1: - return -1 - return max(left, right) + 1 + def isBalanced(self, root: Optional[TreeNode]) -> bool: + return self.get_hight(root) != -1 + def get_hight(self, node): + if not node: + return 0 + left = self.get_hight(node.left) + right = self.get_hight(node.right) + if left == -1 or right == -1 or abs(left - right) > 1: + return -1 + return max(left, right) + 1 ```