From 75e1bb2e7ba1c5334b30068882db8c27cf078982 Mon Sep 17 00:00:00 2001 From: Asterisk <44215173+casnz1601@users.noreply.github.com> Date: Thu, 21 Oct 2021 12:54:24 +0800 Subject: [PATCH 01/35] =?UTF-8?q?Update=200031.=E4=B8=8B=E4=B8=80=E4=B8=AA?= =?UTF-8?q?=E6=8E=92=E5=88=97.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 新增python代码： 抛砖引玉：因题目要求“必须原地修改，只允许使用额外常数空间”，python内置sorted函数以及数组切片+sort()无法使用，故选择另一种算法暂且提供一种python思路 --- problems/0031.下一个排列.md | 27 +++++++++++++++++++++++++++ 1 file changed, 27 insertions(+) diff --git a/problems/0031.下一个排列.md b/problems/0031.下一个排列.md index 9efcb06a..470c354d 100644 --- a/problems/0031.下一个排列.md +++ b/problems/0031.下一个排列.md @@ -122,6 +122,33 @@ class Solution { ## Python ```python +class Solution: + ''' + 抛砖引玉：因题目要求“必须原地修改，只允许使用额外常数空间”，python内置sorted函数以及数组切片+sort()无法使用。 + 故选择另一种算法暂且提供一种python思路 + ''' + def nextPermutation(self, nums: List[int]) -> None: + """ + Do not return anything, modify nums in-place instead. + """ + length = len(nums) + for i in range(length-1, 0, -1): + if nums[i-1] < nums[i]: + for j in range(length-1, 0, -1): + if nums[j] > nums[i-1]: + nums[i-1], nums[j] = nums[j], nums[i-1] + break + self.reverse(nums, i, length-1) + break + else: + # 若正常结束循环，则对原数组直接翻转 + self.reverse(nums, 0, length-1) + + def reverse(self, nums: List[int], low: int, high: int) -> None: + while low < high: + nums[low], nums[high] = nums[high], nums[low] + low += 1 + high -= 1 ``` ## Go From 1598341473568669dae56e3c1e84bef56f090147 Mon Sep 17 00:00:00 2001 From: Asterisk <44215173+casnz1601@users.noreply.github.com> Date: Thu, 21 Oct 2021 12:58:03 +0800 Subject: [PATCH 02/35] =?UTF-8?q?Update=200031.=E4=B8=8B=E4=B8=80=E4=B8=AA?= =?UTF-8?q?=E6=8E=92=E5=88=97.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 补充python sorted()代码 --- problems/0031.下一个排列.md | 17 ++++++++++++++++- 1 file changed, 16 insertions(+), 1 deletion(-) diff --git a/problems/0031.下一个排列.md b/problems/0031.下一个排列.md index 470c354d..9999486e 100644 --- a/problems/0031.下一个排列.md +++ b/problems/0031.下一个排列.md @@ -120,7 +120,22 @@ class Solution { ``` ## Python - +>直接使用sorted()不符合题意 +```python +class Solution: + def nextPermutation(self, nums: List[int]) -> None: + """ + Do not return anything, modify nums in-place instead. + """ + for i in range(len(nums)-1, -1, -1): + for j in range(len(nums)-1, i, -1): + if nums[j] > nums[i]: + nums[j], nums[i] = nums[i], nums[j] + nums[i+1:len(nums)] = sorted(nums[i+1:len(nums)]) + return + nums.sort() +``` +>另一种思路 ```python class Solution: ''' From 99dc8525256fb6f561ce7e47fc711bdac0e87363 Mon Sep 17 00:00:00 2001 From: changlua Date: Thu, 21 Oct 2021 23:54:21 +0800 Subject: [PATCH 03/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=E4=BA=86=E5=89=91?= =?UTF-8?q?=E6=8C=87Offer05.=E6=9B=BF=E6=8D=A2=E7=A9=BA=E6=A0=BC=E7=9A=84j?= =?UTF-8?q?ava=E5=8F=8C=E6=8C=87=E9=92=88=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/剑指Offer05.替换空格.md | 35 ++++++++++++++++++++++++++ 1 file changed, 35 insertions(+) diff --git a/problems/剑指Offer05.替换空格.md b/problems/剑指Offer05.替换空格.md index 232d1a0a..22789450 100644 --- a/problems/剑指Offer05.替换空格.md +++ b/problems/剑指Offer05.替换空格.md @@ -144,6 +144,41 @@ public static String replaceSpace(StringBuffer str) { } return sb.toString(); } + +//方式二：双指针法 +public String replaceSpace(String s) { + if(s == null || s.length() == 0){ + return s; + } + //扩充空间，空格数量2倍 + StringBuilder str = new StringBuilder(); + for (int i = 0; i < s.length(); i++) { + if(s.charAt(i) == ' '){ + str.append(" "); + } + } + //若是没有空格直接返回 + if(str.length() == 0){ + return s; + } + //有空格情况 定义两个指针 + int left = s.length() - 1;//左指针：指向原始字符串最后一个位置 + s += str.toString(); + int right = s.length()-1;//右指针：指向扩展字符串的最后一个位置 + char[] chars = s.toCharArray(); + while(left>=0){ + if(chars[left] == ' '){ + chars[right--] = '0'; + chars[right--] = '2'; + chars[right] = '%'; + }else{ + chars[right] = chars[left]; + } + left--; + right--; + } + return new String(chars); +} ``` From 35154b7be3d47785f77ec3680b71f2201fb2e9c1 Mon Sep 17 00:00:00 2001 From: baici1 <249337001@qq.com> Date: Fri, 22 Oct 2021 16:58:02 +0800 Subject: [PATCH 04/35] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=E4=B9=B0=E5=8D=96?= =?UTF-8?q?=E8=82=A1=E7=A5=A8=E6=9C=80=E4=BD=B3=E6=97=B6=E6=9C=BA=E2=85=A1?= =?UTF-8?q?=EF=BC=8C=E4=B9=B0=E5=8D=96=E8=82=A1=E7=A5=A8=E6=9C=80=E4=BD=B3?= =?UTF-8?q?=E6=97=B6=E6=9C=BA=E2=85=A2=EF=BC=8C=E4=B9=B0=E5=8D=96=E8=82=A1?= =?UTF-8?q?=E7=A5=A8=E6=9C=80=E4=BD=B3=E6=97=B6=E6=9C=BA=E2=85=A3=20go?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...票的最佳时机II（动态规划）.md | 25 +++++++++++++ .../0123.买卖股票的最佳时机III.md | 32 +++++++++++++++++ .../0188.买卖股票的最佳时机IV.md | 36 +++++++++++++++++++ 3 files changed, 93 insertions(+) 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 4e64beaf..30c56c25 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 @@ -200,6 +200,31 @@ class Solution: Go： +```go +func maxProfit(prices []int) int { + //创建数组 + dp:=make([][]int,len(prices)) + for i:=0;ib{ + return a + } + return b +} +``` + + + JavaScript： > 版本一： 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 5cffbe9e..a97f293a 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 @@ -276,6 +276,42 @@ class Solution: ``` Go： +```go +func maxProfit(k int, prices []int) int { + if len(prices)==0{ + return 0 + } + dp:=make([][]int,len(prices)) + for i:=0;ib{ + return a + } + return b +} +``` + + + Javascript： From 5a69916e0ff8c569b85090b6af97d9bddefff149 Mon Sep 17 00:00:00 2001 From: mengyuan Date: Fri, 22 Oct 2021 17:26:47 +0800 Subject: [PATCH 05/35] =?UTF-8?q?update:=200202.=E5=BF=AB=E4=B9=90?= =?UTF-8?q?=E6=95=B0=20js=E7=89=88=E6=9C=AC=20=E4=BD=BF=E7=94=A8Set()?= =?UTF-8?q?=EF=BC=8C=E6=B1=82=E5=92=8C=E7=94=A8reduce?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0202.快乐数.md | 17 +++++++++++++++++ 1 file changed, 17 insertions(+) diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md index 5704283d..d7a6b4e9 100644 --- a/problems/0202.快乐数.md +++ b/problems/0202.快乐数.md @@ -215,6 +215,23 @@ var isHappy = function(n) { } return n === 1; }; + +// 方法四：使用Set()，求和用reduce +var isHappy = function(n) { + let set = new Set(); + let totalCount; + while(totalCount !== 1) { + let arr = (''+(totalCount || n)).split(''); + totalCount = arr.reduce((total, num) => { + return total + num * num + }, 0) + if (set.has(totalCount)) { + return false; + } + set.add(totalCount); + } + return true; +}; ``` Swift： From 344dce02b36b6af546ccbe308c2e599f2b0312f0 Mon Sep 17 00:00:00 2001 From: Junwen Huang Date: Sat, 23 Oct 2021 10:32:35 +0800 Subject: [PATCH 06/35] =?UTF-8?q?Update=200257.=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=E7=9A=84=E6=89=80=E6=9C=89=E8=B7=AF=E5=BE=84.md=20=E2=80=94?= =?UTF-8?q?=E2=80=94=20=E5=A2=9E=E5=8A=A0go=E4=B8=8Ejs=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E7=9A=84=E9=80=92=E5=BD=92=E8=A7=A3=E6=B3=95?= 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 | 66 ++++++++++++++++++++++- 1 file changed, 65 insertions(+), 1 deletion(-) 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 f902aab2..b15fa1be 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 @@ -469,6 +469,9 @@ class Solution: --- Go： + +递归法： + ```go func binaryTreePaths(root *TreeNode) []string { res := make([]string, 0) @@ -491,10 +494,45 @@ func binaryTreePaths(root *TreeNode) []string { return res } ``` + +迭代法： + +```go +func binaryTreePaths(root *TreeNode) []string { + stack := []*TreeNode{} + paths := make([]string, 0) + res := make([]string, 0) + if root != nil { + stack = append(stack, root) + paths = append(paths, "") + } + for len(stack) > 0 { + l := len(stack) + node := stack[l-1] + path := paths[l-1] + stack = stack[:l-1] + paths = paths[:l-1] + if node.Left == nil && node.Right == nil { + res = append(res, path+strconv.Itoa(node.Val)) + continue + } + if node.Right != nil { + stack = append(stack, node.Right) + paths = append(paths, path+strconv.Itoa(node.Val)+"->") + } + if node.Left != nil { + stack = append(stack, node.Left) + paths = append(paths, path+strconv.Itoa(node.Val)+"->") + } + } + return res +} +``` + --- JavaScript: -1.递归版本 +递归法： ```javascript var binaryTreePaths = function(root) { @@ -518,6 +556,32 @@ var binaryTreePaths = function(root) { }; ``` +迭代法： + +```javascript +var binaryTreePaths = function(root) { + if (!root) return []; + const stack = [root], paths = [''], res = []; + while (stack.length) { + const node = stack.pop(); + let path = paths.pop(); + if (!node.left && !node.right) { // 到叶子节点终止, 添加路径到结果中 + res.push(path + node.val); + continue; + } + path += node.val + '->'; + if (node.right) { // 右节点存在 + stack.push(node.right); + paths.push(path); + } + if (node.left) { // 左节点存在 + stack.push(node.left); + paths.push(path); + } + } + return res; +}; +``` From fdca5f00ffd6108b544ff6d647fd88fb7490dd54 Mon Sep 17 00:00:00 2001 From: SambacFeng <56753082+SambacFeng@users.noreply.github.com> Date: Sat, 23 Oct 2021 16:32:28 +0800 Subject: [PATCH 07/35] =?UTF-8?q?update=200450.=E5=88=A0=E9=99=A4=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84=E8=8A=82?= =?UTF-8?q?=E7=82=B9=EF=BC=8C=E5=A2=9E=E5=8A=A0js=E7=9A=84=E8=BF=AD?= =?UTF-8?q?=E4=BB=A3=E5=86=99=E6=B3=95?= 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 | 35 +++++++++++++++++++ 1 file changed, 35 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 32dd862e..23e710ad 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 @@ -485,6 +485,41 @@ var deleteNode = function (root, key) { }; ``` +迭代 +``` javascript +var deleteNode = function (root, key) { + const deleteOneNode = target => { + if (!target) return target + if (!target.right) return target.left + let cur = target.right + while (cur.left) { + cur = cur.left + } + cur.left = target.left + return target.right + } + + if (!root) return root + let cur = root + let pre = null + while (cur) { + if (cur.val === key) break + pre = cur + cur.val > key ? cur = cur.left : cur = cur.right + } + if (!pre) { + return deleteOneNode(cur) + } + if (pre.left && pre.left.val === key) { + pre.left = deleteOneNode(cur) + } + if (pre.right && pre.right.val === key) { + pre.right = deleteOneNode(cur) + } + return root +} +``` + ----------------------- From 74cb27fa2c8106fbcd6a25344ae474f81078f198 Mon Sep 17 00:00:00 2001 From: Junwen Huang Date: Sat, 23 Oct 2021 20:47:43 +0800 Subject: [PATCH 08/35] =?UTF-8?q?Update=200106.=E4=BB=8E=E4=B8=AD=E5=BA=8F?= =?UTF-8?q?=E4=B8=8E=E5=90=8E=E5=BA=8F=E9=81=8D=E5=8E=86=E5=BA=8F=E5=88=97?= =?UTF-8?q?=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 优化js版本代码并添加注释说明 --- ...序与后序遍历序列构造二叉树.md | 31 +++++++++---------- 1 file changed, 14 insertions(+), 17 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 a83341fd..11083a59 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 @@ -792,16 +792,13 @@ func findRootIndex(target int,inorder []int) int{ ```javascript var buildTree = function(inorder, postorder) { - if (!postorder.length) return null - - let root = new TreeNode(postorder[postorder.length - 1]) - - let index = inorder.findIndex(number => number === root.val) - - root.left = buildTree(inorder.slice(0, index), postorder.slice(0, index)) - root.right = buildTree(inorder.slice(index + 1, inorder.length), postorder.slice(index, postorder.length - 1)) - - return root + if (!preorder.length) return null; + const rootVal = postorder.pop(); // 从后序遍历的数组中获取中间节点的值， 即数组最后一个值 + let rootIndex = inorder.indexOf(rootVal); // 获取中间节点在中序遍历中的下标 + const root = new TreeNode(rootVal); // 创建中间节点 + root.left = buildTree(inorder.slice(0, rootIndex), postorder.slice(0, rootIndex)); // 创建左节点 + root.right = buildTree(inorder.slice(rootIndex + 1), postorder.slice(rootIndex)); // 创建右节点 + return root; }; ``` @@ -809,13 +806,13 @@ var buildTree = function(inorder, postorder) { ```javascript var buildTree = function(preorder, inorder) { - if(!preorder.length) - return null; - let root = new TreeNode(preorder[0]); - let mid = inorder.findIndex((number) => number === root.val); - root.left = buildTree(preorder.slice(1, mid + 1), inorder.slice(0, mid)); - root.right = buildTree(preorder.slice(mid + 1, preorder.length), inorder.slice(mid + 1, inorder.length)); - return root; + if (!preorder.length) return null; + const rootVal = preorder.shift(); // 从前序遍历的数组中获取中间节点的值， 即数组第一个值 + const index = inorder.indexOf(rootVal); // 获取中间节点在中序遍历中的下标 + const root = new TreeNode(rootVal); // 创建中间节点 + root.left = buildTree(preorder.slice(0, index), inorder.slice(0, index)); // 创建左节点 + root.right = buildTree(preorder.slice(index), inorder.slice(index + 1)); // 创建右节点 + return root; }; ``` From 90ed55b5da370129f9d50691c7d1985642c2a818 Mon Sep 17 00:00:00 2001 From: hailincai Date: Sun, 24 Oct 2021 08:01:53 -0400 Subject: [PATCH 09/35] =?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 修改错别字 --- problems/二叉树的递归遍历.md | 2 +- 1 file changed, 1 insertion(+), 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 eee150b7..209cbcb7 100644 --- a/problems/二叉树的递归遍历.md +++ b/problems/二叉树的递归遍历.md @@ -54,7 +54,7 @@ traversal(cur->left, vec); // 左 traversal(cur->right, vec); // 右 ``` -单层递归的逻辑就是按照中左右的顺序来处理的，这样二叉树的前序遍历，基本就写完了，在看一下完整代码： +单层递归的逻辑就是按照中左右的顺序来处理的，这样二叉树的前序遍历，基本就写完了，再看一下完整代码： 前序遍历： From c8cc5b51207e045f594209332009d985545ca52c Mon Sep 17 00:00:00 2001 From: hailincai Date: Sun, 24 Oct 2021 09:51:56 -0400 Subject: [PATCH 10/35] =?UTF-8?q?Update=200053.=E6=9C=80=E5=A4=A7=E5=AD=90?= =?UTF-8?q?=E5=BA=8F=E5=92=8C.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 增加dp方法 --- problems/0053.最大子序和.md | 19 +++++++++++++++++++ 1 file changed, 19 insertions(+) diff --git a/problems/0053.最大子序和.md b/problems/0053.最大子序和.md index 75281210..5c45aa0a 100644 --- a/problems/0053.最大子序和.md +++ b/problems/0053.最大子序和.md @@ -161,6 +161,25 @@ class Solution { } ``` +```java +// DP 方法 +class Solution { + public int maxSubArray(int[] nums) { + int ans = Integer.MIN_VALUE; + int[] dp = new int[nums.length]; + dp[0] = nums[0]; + ans = dp[0]; + + for (int i = 1; i < nums.length; i++){ + dp[i] = Math.max(dp[i-1] + nums[i], nums[i]); + ans = Math.max(dp[i], ans); + } + + return ans; + } +} +``` + Python： ```python class Solution: From f9abccf8892f4d7054806a97a84110846e3d047f Mon Sep 17 00:00:00 2001 From: hailincai Date: Sun, 24 Oct 2021 09:53:42 -0400 Subject: [PATCH 11/35] =?UTF-8?q?Update=200376.=E6=91=86=E5=8A=A8=E5=BA=8F?= =?UTF-8?q?=E5=88=97.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 增加dp方法 --- problems/0376.摆动序列.md | 30 ++++++++++++++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/problems/0376.摆动序列.md b/problems/0376.摆动序列.md index 5587a8c7..e58a26ff 100644 --- a/problems/0376.摆动序列.md +++ b/problems/0376.摆动序列.md @@ -199,6 +199,36 @@ class Solution { } ``` +```java +// DP +class Solution { + public int wiggleMaxLength(int[] nums) { + // 0 i 作为波峰的最大长度 + // 1 i 作为波谷的最大长度 + int dp[][] = new int[nums.length][2]; + + dp[0][0] = dp[0][1] = 1; + for (int i = 1; i < nums.length; i++){ + //i 自己可以成为波峰或者波谷 + dp[i][0] = dp[i][1] = 1; + + for (int j = 0; j < i; j++){ + if (nums[j] > nums[i]){ + // i 是波谷 + dp[i][1] = Math.max(dp[i][1], dp[j][0] + 1); + } + if (nums[j] < nums[i]){ + // i 是波峰 + dp[i][0] = Math.max(dp[i][0], dp[j][1] + 1); + } + } + } + + return Math.max(dp[nums.length - 1][0], dp[nums.length - 1][1]); + } +} +``` + Python： ```python3 class Solution: From 61838527a072acd7cf7fb4e3fce8de0f4ff91fce Mon Sep 17 00:00:00 2001 From: hailincai Date: Sun, 24 Oct 2021 10:07:34 -0400 Subject: [PATCH 12/35] =?UTF-8?q?Update=200509.=E6=96=90=E6=B3=A2=E9=82=A3?= =?UTF-8?q?=E5=A5=91=E6=95=B0.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 增加java非压缩状态版本，易于理解 --- problems/0509.斐波那契数.md | 20 ++++++++++++++++++++ 1 file changed, 20 insertions(+) diff --git a/problems/0509.斐波那契数.md b/problems/0509.斐波那契数.md index 4102dd64..cb54a0f9 100644 --- a/problems/0509.斐波那契数.md +++ b/problems/0509.斐波那契数.md @@ -186,6 +186,26 @@ class Solution { } ``` +```java +//非压缩状态的版本 +class Solution { + public int fib(int n) { + if (n <= 1) return n; + + int[] dp = new int[n + 1]; + + dp[0] = 0; + dp[1] = 1; + + for (int index = 2; index <= n; index++){ + dp[index] = dp[index - 1] + dp[index - 2]; + } + + return dp[n]; + } +} +``` + Python： ```python3 class Solution: From 09e82fa59602210006e8fbc9977119bcd3b3030c Mon Sep 17 00:00:00 2001 From: hailincai Date: Sun, 24 Oct 2021 10:28:31 -0400 Subject: [PATCH 13/35] =?UTF-8?q?Update=20=E5=93=88=E5=B8=8C=E8=A1=A8?= =?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 fix typo --- problems/哈希表理论基础.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/哈希表理论基础.md b/problems/哈希表理论基础.md index aeb11b2e..05fb2a32 100644 --- a/problems/哈希表理论基础.md +++ b/problems/哈希表理论基础.md @@ -32,7 +32,7 @@ ## 哈希函数 -哈希函数，把学生的姓名直接映射为哈希表上的索引，然后就可以通过查询索引下表快速知道这位同学是否在这所学校里了。 +哈希函数，把学生的姓名直接映射为哈希表上的索引，然后就可以通过查询索引下标快速知道这位同学是否在这所学校里了。 哈希函数如下图所示，通过hashCode把名字转化为数值，一般hashcode是通过特定编码方式，可以将其他数据格式转化为不同的数值，这样就把学生名字映射为哈希表上的索引数字了。 From 825600a1d2c67f34fd9ecae91a171475c1ee1cb7 Mon Sep 17 00:00:00 2001 From: sugarlesss Date: Mon, 25 Oct 2021 10:38:34 +0800 Subject: [PATCH 14/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=20=E9=9D=A2=E8=AF=95?= =?UTF-8?q?=E9=A2=9802.07.=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4=20=E9=94=99?= =?UTF-8?q?=E5=88=AB=E5=AD=97=E5=8B=98=E8=AF=AF?= 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 | 2 +- 1 file changed, 1 insertion(+), 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 f283b161..c5b8ed6f 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 @@ -38,7 +38,7 @@ ![面试题02.07.链表相交_2](https://code-thinking.cdn.bcebos.com/pics/面试题02.07.链表相交_2.png) -此时我们就可以比较curA和curB是否相同，如果不相同，同时向后移动curA和curB，如果遇到curA == curB，则找到焦点。 +此时我们就可以比较curA和curB是否相同，如果不相同，同时向后移动curA和curB，如果遇到curA == curB，则找到交点。 否则循环退出返回空指针。 From 28e3d95aa9b3e7bace52ec57a018302c22a7e9f8 Mon Sep 17 00:00:00 2001 From: Wen Date: Tue, 26 Oct 2021 22:03:21 +0800 Subject: [PATCH 15/35] =?UTF-8?q?=E6=96=B0=E5=A2=9E=200213.=E6=89=93?= =?UTF-8?q?=E5=AE=B6=E5=8A=AB=E8=88=8DII.md=20Go=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0213.打家劫舍II.md | 34 +++++++++++++++++++++++++++++++++ 1 file changed, 34 insertions(+) diff --git a/problems/0213.打家劫舍II.md b/problems/0213.打家劫舍II.md index 332d3218..4c536f64 100644 --- a/problems/0213.打家劫舍II.md +++ b/problems/0213.打家劫舍II.md @@ -166,7 +166,41 @@ const robRange = (nums, start, end) => { } ``` Go： +```go +// 打家劫舍Ⅱ 动态规划 +// 时间复杂度O(n) 空间复杂度O(n) +func rob(nums []int) int { + if len(nums) == 1 { + return nums[0] + } + if len(nums) == 2 { + return max(nums[0], nums[1]) + } + + result1 := robRange(nums, 0) + result2 := robRange(nums, 1) + return max(result1, result2) +} +// 偷盗指定的范围 +func robRange(nums []int, start int) int { + dp := make([]int, len(nums)) + dp[1] = nums[start] + + for i := 2; i < len(nums); i++ { + dp[i] = max(dp[i - 2] + nums[i - 1 + start], dp[i - 1]) + } + + return dp[len(nums) - 1] +} + +func max(a, b int) int { + if a > b { + return a + } + return b +} +``` From 9b793937638a23bc69968c8e93816ee682712ca0 Mon Sep 17 00:00:00 2001 From: Wen Date: Tue, 26 Oct 2021 22:05:23 +0800 Subject: [PATCH 16/35] =?UTF-8?q?=E6=96=B0=E5=A2=9E=200337.=E6=89=93?= =?UTF-8?q?=E5=AE=B6=E5=8A=AB=E8=88=8DIII.md=20Go=E8=A7=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0337.打家劫舍III.md | 29 +++++++++++++++++++++++++++++ 1 file changed, 29 insertions(+) diff --git a/problems/0337.打家劫舍III.md b/problems/0337.打家劫舍III.md index dfb8ba57..541c1163 100644 --- a/problems/0337.打家劫舍III.md +++ b/problems/0337.打家劫舍III.md @@ -395,6 +395,35 @@ const rob = root => { }; ``` +Go: +```go +// 打家劫舍Ⅲ 动态规划 +// 时间复杂度O(n) 空间复杂度O(logn) +func rob(root *TreeNode) int { + dp := traversal(root) + return max(dp[0], dp[1]) +} + +func traversal(cur *TreeNode) []int { + if cur == nil { + return []int{0, 0} + } + + dpL := traversal(cur.Left) + dpR := traversal(cur.Right) + + val1 := cur.Val + dpL[0] + dpR[0] // 偷盗当前节点 + val2 := max(dpL[0], dpL[1]) + max(dpR[0], dpR[1]) // 不偷盗当前节点 + return []int{val2, val1} +} + +func max(a, b int) int { + if a > b { + return a + } + return b +} +``` From c5a9eadab17dcea7e9d693f144c7f2669d5f4c54 Mon Sep 17 00:00:00 2001 From: Arthur Date: Tue, 26 Oct 2021 15:09:19 +0100 Subject: [PATCH 17/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200108.=E5=B0=86?= =?UTF-8?q?=E6=9C=89=E5=BA=8F=E6=95=B0=E7=BB=84=E8=BD=AC=E6=8D=A2=E4=B8=BA?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E7=89=88=E6=9C=AC?= 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 | 19 ++++++++++++++++++- 1 file changed, 18 insertions(+), 1 deletion(-) 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 e82d6e95..7dcce649 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 @@ -375,8 +375,25 @@ var sortedArrayToBST = function (nums) { }; ``` +##C +递归 +```c +struct TreeNode* traversal(int* nums, int left, int right) { + if (left > right) + return NULL; + int mid = left + ((right - left) / 2); + struct TreeNode* root = (struct TreeNode*)malloc(sizeof(struct TreeNode)); + root->val = nums[mid]; + root->left = traversal(nums, left, mid - 1); + root->right = traversal(nums, mid + 1, right); + return root; +} - +struct TreeNode* sortedArrayToBST(int* nums, int numsSize) { + struct TreeNode* root = traversal(nums, 0, numsSize - 1); + return root; +} +``` ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) From 79623eba4afb9700fed82d3b0a6d44fa7870e98e Mon Sep 17 00:00:00 2001 From: Arthur Date: Tue, 26 Oct 2021 15:59:35 +0100 Subject: [PATCH 18/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200538.=E6=8A=8A?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91=E8=BD=AC=E6=8D=A2?= =?UTF-8?q?=E4=B8=BA=E7=B4=AF=E5=8A=A0=E6=A0=91.md=20C=E8=AF=AD=E8=A8=80?= 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 | 21 +++++++++++++++++++ 1 file changed, 21 insertions(+) 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 6de98c6f..16b00817 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 @@ -293,6 +293,27 @@ var convertBST = function (root) { }; ``` +##C + +递归 +```c +int pre; +void traversal(struct TreeNode* node) { + if(!node) + return ; + traversal(node->right); + node->val = node->val + pre; + pre = node->val; + traversal(node->left); +} + +struct TreeNode* convertBST(struct TreeNode* root){ + pre = 0; + traversal(root); + return root; +} +``` + ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) * B站视频：[代码随想录](https://space.bilibili.com/525438321) From 5eab2050cf9c1d419491f085796d7e788dfac317 Mon Sep 17 00:00:00 2001 From: Younglesszzz <571688981@qq.com> Date: Wed, 27 Oct 2021 20:03:18 +0800 Subject: [PATCH 19/35] =?UTF-8?q?=E5=8F=A6=E4=B8=80=E7=A7=8D=E5=86=99?= =?UTF-8?q?=E6=B3=95=20=E5=85=A8=E6=8E=92=E5=88=97=E2=85=A1=20Java?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...溯算法去重问题的另一种写法.md | 39 +++++++++++++++++++ 1 file changed, 39 insertions(+) diff --git a/problems/回溯算法去重问题的另一种写法.md b/problems/回溯算法去重问题的另一种写法.md index d8125e91..8f5744b5 100644 --- a/problems/回溯算法去重问题的另一种写法.md +++ b/problems/回溯算法去重问题的另一种写法.md @@ -249,6 +249,45 @@ used数组可是全局变量，每层与每层之间公用一个used数组，所 Java： +**47.全排列II** + + +```java +class Solution { + private List> res = new ArrayList<>(); + private List path = new ArrayList<>(); + private boolean[] used = null; + + public List> permuteUnique(int[] nums) { + used = new boolean[nums.length]; + Arrays.sort(nums); + backtracking(nums); + return res; + } + + public void backtracking(int[] nums) { + if (path.size() == nums.length) { + res.add(new ArrayList<>(path)); + return; + } + HashSet hashSet = new HashSet<>();//层去重 + for (int i = 0; i < nums.length; i++) { + if (hashSet.contains(nums[i])) + continue; + if (used[i] == true)//枝去重 + continue; + hashSet.add(nums[i]);//记录元素 + used[i] = true; + path.add(nums[i]); + backtracking(nums); + path.remove(path.size() - 1); + used[i] = false; + } + } +} + +``` + Python： From ff44674da3b09b87fbfcf33d20037e1d182c9baa Mon Sep 17 00:00:00 2001 From: changlua Date: Wed, 27 Oct 2021 23:49:15 +0800 Subject: [PATCH 20/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=E4=BA=860151.=E7=BF=BB?= =?UTF-8?q?=E8=BD=AC=E5=AD=97=E7=AC=A6=E4=B8=B2=E9=87=8C=E7=9A=84=E5=8D=95?= =?UTF-8?q?=E8=AF=8DJava=E4=B8=A4=E7=A7=8D=E8=A7=A3=E6=B3=95=E2=91=A0?= =?UTF-8?q?=E5=AD=97=E7=AC=A6=E6=95=B0=E7=BB=84=E5=A1=AB=E5=85=85=E2=91=A1?= =?UTF-8?q?=E5=8F=8C=E5=8F=8D=E8=BD=AC+=E7=A7=BB=E4=BD=8D?= 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 | 88 ++++++++++++++++++++ 1 file changed, 88 insertions(+) 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 8d28b332..bf274263 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 @@ -300,6 +300,94 @@ class Solution { } ``` +```java +//解法二：创建新字符数组填充。时间复杂度O(n) +class Solution { + public String reverseWords(String s) { + //源字符数组 + char[] initialArr = s.toCharArray(); + //新字符数组 + char[] newArr = new char[initialArr.length+1];//下面循环添加"单词 "，最终末尾的空格不会返回 + int newArrPos = 0; + //i来进行整体对源字符数组从后往前遍历 + int i = initialArr.length-1; + while(i>=0){ + while(i>=0 && initialArr[i] == ' '){i--;} //跳过空格 + //此时i位置是边界或!=空格，先记录当前索引，之后的while用来确定单词的首字母的位置 + int right = i; + while(i>=0 && initialArr[i] != ' '){i--;} + //指定区间单词取出(由于i为首字母的前一位，所以这里+1,)，取出的每组末尾都带有一个空格 + for (int j = i+1; j <= right; j++) { + newArr[newArrPos++] = initialArr[j]; + if(j == right){ + newArr[newArrPos++] = ' ';//空格 + } + } + } + //若是原始字符串没有单词，直接返回空字符串；若是有单词，返回0-末尾空格索引前范围的字符数组(转成String返回) + if(newArrPos == 0){ + return ""; + }else{ + return new String(newArr,0,newArrPos-1); + } + } +} +``` + +```java +//解法三：双反转+移位，在原始数组上进行反转。空间复杂度O(1) +class Solution { + /** + * 思路： + * ①反转字符串 "the sky is blue " => " eulb si yks eht" + * ②遍历 " eulb si yks eht"，每次先对某个单词进行反转再移位 + * 这里以第一个单词进行为演示：" eulb si yks eht" ==反转=> " blue si yks eht" ==移位=> "blue si yks eht" + */ + public String reverseWords(String s) { + //步骤1：字符串整体反转（此时其中的单词也都反转了） + char[] initialArr = s.toCharArray(); + reverse(initialArr, 0, s.length() - 1); + int k = 0; + for (int i = 0; i < initialArr.length; i++) { + if (initialArr[i] == ' ') { + continue; + } + int tempCur = i; + while (i < initialArr.length && initialArr[i] != ' ') { + i++; + } + for (int j = tempCur; j < i; j++) { + if (j == tempCur) { //步骤二：二次反转 + reverse(initialArr, tempCur, i - 1);//对指定范围字符串进行反转，不反转从后往前遍历一个个填充有问题 + } + //步骤三：移动操作 + initialArr[k++] = initialArr[j]; + if (j == i - 1) { //遍历结束 + //避免越界情况，例如=> "asdasd df f"，不加判断最后就会数组越界 + if (k < initialArr.length) { + initialArr[k++] = ' '; + } + } + } + } + if (k == 0) { + return ""; + } else { + //参数三：以防出现如"asdasd df f"=>"f df asdasd"正好凑满不需要省略空格情况 + return new String(initialArr, 0, (k == initialArr.length) && (initialArr[k - 1] != ' ') ? k : k - 1); + } + } + + public void reverse(char[] chars, int begin, int end) { + for (int i = begin, j = end; i < j; i++, j--) { + chars[i] ^= chars[j]; + chars[j] ^= chars[i]; + chars[i] ^= chars[j]; + } + } +} +``` + python: ```Python From 870fdf9824fd92ad82672b14c51a20117d42dbd2 Mon Sep 17 00:00:00 2001 From: changfubai Date: Thu, 28 Oct 2021 17:52:04 +0800 Subject: [PATCH 21/35] =?UTF-8?q?=E8=B0=83=E6=95=B4=E7=AC=94=E8=AF=AF<2020?= =?UTF-8?q?1224=E8=B4=AA=E5=BF=83=E5=91=A8=E6=9C=AB=E6=80=BB=E7=BB=93>?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/周总结/20201224贪心周末总结.md | 8 ++++---- 1 file changed, 4 insertions(+), 4 deletions(-) diff --git a/problems/周总结/20201224贪心周末总结.md b/problems/周总结/20201224贪心周末总结.md index 097ae9ed..1bc39247 100644 --- a/problems/周总结/20201224贪心周末总结.md +++ b/problems/周总结/20201224贪心周末总结.md @@ -6,7 +6,7 @@ 在[贪心算法：用最少数量的箭引爆气球](https://programmercarl.com/0452.用最少数量的箭引爆气球.html)中，我们开始讲解了重叠区间问题，用最少的弓箭射爆所有气球，其本质就是找到最大的重叠区间。 -按照左边界经行排序后，如果气球重叠了，重叠气球中右边边界的最小值 之前的区间一定需要一个弓箭 +按照左边界进行排序后，如果气球重叠了，重叠气球中右边边界的最小值 之前的区间一定需要一个弓箭 如图： @@ -28,7 +28,7 @@ 弓箭的数量就相当于是非交叉区间的数量，只要把弓箭那道题目代码里射爆气球的判断条件加个等号（认为[0，1][1，2]不是相邻区间），然后用总区间数减去弓箭数量 就是要移除的区间数量了。 -把[贪心算法：用最少数量的箭引爆气球](https://programmercarl.com/0452.用最少数量的箭引爆气球.html)代码稍做修改，别可以AC本题。 +把[贪心算法：用最少数量的箭引爆气球](https://programmercarl.com/0452.用最少数量的箭引爆气球.html)代码稍做修改，就可以AC本题。 修改后的C++代码如下： ```CPP @@ -60,7 +60,7 @@ public: [贪心算法：划分字母区间](https://programmercarl.com/0763.划分字母区间.html)中我们要把这个字符串划分为尽可能多的片段，同一字母最多出现在一个片段中。 -这道题目leetcode上标的是贪心，其实我不认识是贪心，因为没感受到局部最优和全局最优的关系。 +这道题目leetcode上标的是贪心，其实我不认为是贪心，因为没感受到局部最优和全局最优的关系。 但不影响这是一道好题，思路很不错，**通过字符出现最远距离取并集的方法，把出现过的字符都圈到一个区间里**。 @@ -91,7 +91,7 @@ public: ## 总结 -本周的主题就是用贪心算法来解决区间问题，进过本周的学习，大家应该对区间的各种合并分割有一定程度的了解了。 +本周的主题就是用贪心算法来解决区间问题，经过本周的学习，大家应该对区间的各种合并分割有一定程度的了解了。 其实很多区间的合并操作看起来都是常识，其实贪心算法有时候就是常识，哈哈，但也别小看了贪心算法。 From c9845b1153addce3d5cc95ecb2864f22f5af16b7 Mon Sep 17 00:00:00 2001 From: Arthur Date: Thu, 28 Oct 2021 12:22:05 +0100 Subject: [PATCH 22/35] =?UTF-8?q?=E4=BF=AE=E6=94=B9=200108.=E5=B0=86?= =?UTF-8?q?=E6=9C=89=E5=BA=8F=E6=95=B0=E7=BB=84=E8=BD=AC=E6=8D=A2=E4=B8=BA?= =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=90=9C=E7=B4=A2=E6=A0=91.md=20=E6=B3=A8?= =?UTF-8?q?=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0108.将有序数组转换为二叉搜索树.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) 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 7dcce649..213eb2b9 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 @@ -375,7 +375,7 @@ var sortedArrayToBST = function (nums) { }; ``` -##C +## C 递归 ```c struct TreeNode* traversal(int* nums, int left, int right) { From a8ef4fe6eeea4869dac2361566e3ed11e3d8d6a6 Mon Sep 17 00:00:00 2001 From: hailincai Date: Fri, 29 Oct 2021 08:16:18 -0400 Subject: [PATCH 23/35] =?UTF-8?q?Update=200104.=E4=BA=8C=E5=8F=89=E6=A0=91?= =?UTF-8?q?=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 增加559N Java 递归方法 --- problems/0104.二叉树的最大深度.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) 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 e20f147f..ff7cbfd1 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 @@ -312,6 +312,24 @@ class solution { ``` ### 559.n叉树的最大深度 +```java +class Solution { + /*递归法，后序遍历求root节点的高度*/ + public int maxDepth(Node root) { + if (root == null) return 0; + + int depth = 0; + if (root.children != null){ + for (Node child : root.children){ + depth = Math.max(depth, maxDepth(child)); + } + } + + return depth + 1; //中节点 + } +} +``` + ```java class solution { /** From 50b63f844131e703e2191cfa0442394079092708 Mon Sep 17 00:00:00 2001 From: konng Date: Sun, 31 Oct 2021 13:19:47 +0800 Subject: [PATCH 24/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=20golang=20=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91=E9=81=8D=E5=8E=86=E7=9A=84=E6=89=80=E6=9C=89?= =?UTF-8?q?=E4=BB=A3=E7=A0=81=EF=BC=8C=E7=BD=AE=E6=8D=A2=E4=B8=BA=E5=92=8C?= =?UTF-8?q?=20c++=20=E6=A8=A1=E6=9D=BF=E7=9B=B8=E5=90=8C=E7=9A=84=E9=80=BB?= =?UTF-8?q?=E8=BE=91=EF=BC=8C=E8=80=8C=E4=B8=8D=E6=98=AF=E6=96=B0=E7=9A=84?= =?UTF-8?q?=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 之前的 golang 代码很不美观，并且他的逻辑和 c++ 的不一样，谁还专门记两个模板啊， 并且连缩进也没有，所以我就重写了一份 现在的 golang 逻辑和 c++ 相同 并且也有缩进，也没有废话代码，舒服多了 --- problems/二叉树的迭代遍历.md | 132 ++++++++++++--------------- 1 file changed, 60 insertions(+), 72 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 6889ecc0..7aff0541 100644 --- a/problems/二叉树的迭代遍历.md +++ b/problems/二叉树的迭代遍历.md @@ -299,108 +299,96 @@ class Solution: Go： + > 迭代法前序遍历 ```go -//迭代法前序遍历 -/** - type Element struct { - // 元素保管的值 - Value interface{} - // 内含隐藏或非导出字段 -} - -func (l *List) Back() *Element -前序遍历：中左右 -压栈顺序：右左中 - **/ func preorderTraversal(root *TreeNode) []int { + ans := []int{} + if root == nil { - return nil + return ans } - var stack = list.New() - stack.PushBack(root.Right) - stack.PushBack(root.Left) - res:=[]int{} - res=append(res,root.Val) - for stack.Len()>0 { - e:=stack.Back() - stack.Remove(e) - node := e.Value.(*TreeNode)//e是Element类型，其值为e.Value.由于Value为接口，所以要断言 - if node==nil{ - continue + + st := list.New() + st.PushBack(root) + + for st.Len() > 0 { + node := st.Remove(st.Back()).(*TreeNode) + + ans = append(ans, node.Val) + if node.Right != nil { + st.PushBack(node.Right) + } + if node.Left != nil { + st.PushBack(node.Left) } - res=append(res,node.Val) - stack.PushBack(node.Right) - stack.PushBack(node.Left) } - return res + return ans } ``` > 迭代法后序遍历 ```go -//迭代法后序遍历 -//后续遍历：左右中 -//压栈顺序：中右左(按照前序遍历思路)，再反转结果数组 func postorderTraversal(root *TreeNode) []int { + ans := []int{} + if root == nil { - return nil + return ans } - var stack = list.New() - stack.PushBack(root.Left) - stack.PushBack(root.Right) - res:=[]int{} - res=append(res,root.Val) - for stack.Len()>0 { - e:=stack.Back() - stack.Remove(e) - node := e.Value.(*TreeNode)//e是Element类型，其值为e.Value.由于Value为接口，所以要断言 - if node==nil{ - continue + + st := list.New() + st.PushBack(root) + + for st.Len() > 0 { + node := st.Remove(st.Back()).(*TreeNode) + + ans = append(ans, node.Val) + if node.Left != nil { + st.PushBack(node.Left) + } + if node.Right != nil { + st.PushBack(node.Right) } - res=append(res,node.Val) - stack.PushBack(node.Left) - stack.PushBack(node.Right) } - for i:=0;i 迭代法中序遍历 ```go -//迭代法中序遍历 func inorderTraversal(root *TreeNode) []int { - rootRes:=[]int{} - if root==nil{ - return nil + ans := []int{} + if root == nil { + return ans } - stack:=list.New() - node:=root - //先将所有左节点找到，加入栈中 - for node!=nil{ - stack.PushBack(node) - node=node.Left - } - //其次对栈中的每个节点先弹出加入到结果集中，再找到该节点的右节点的所有左节点加入栈中 - for stack.Len()>0{ - e:=stack.Back() - node:=e.Value.(*TreeNode) - stack.Remove(e) - //找到该节点的右节点，再搜索他的所有左节点加入栈中 - rootRes=append(rootRes,node.Val) - node=node.Right - for node!=nil{ - stack.PushBack(node) - node=node.Left + + st := list.New() + cur := root + + for cur != nil || st.Len() > 0 { + if cur != nil { + st.PushBack(cur) + cur = cur.Left + } else { + cur = st.Remove(st.Back()).(*TreeNode) + ans = append(ans, cur.Val) + cur = cur.Right } } - return rootRes + + return ans } ``` From b87d3266b6604aaaac139b03f7e28e3943d92f5f Mon Sep 17 00:00:00 2001 From: CJ-cooper6 <71270517+CJ-cooper6@users.noreply.github.com> Date: Sun, 31 Oct 2021 16:37:04 +0800 Subject: [PATCH 25/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0225.=20=E7=94=A8?= =?UTF-8?q?=E9=98=9F=E5=88=97=E5=AE=9E=E7=8E=B0=E6=A0=88=20=20Go=20?= =?UTF-8?q?=E4=B8=A4=E4=B8=AA=E9=98=9F=E5=88=97=E5=AE=9E=E7=8E=B0=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 添加225. 用队列实现栈 Go 两个队列实现版本 --- problems/0225.用队列实现栈.md | 66 +++++++++++++++++++++++++++++ 1 file changed, 66 insertions(+) diff --git a/problems/0225.用队列实现栈.md b/problems/0225.用队列实现栈.md index d9819626..c3a6ef8e 100644 --- a/problems/0225.用队列实现栈.md +++ b/problems/0225.用队列实现栈.md @@ -359,6 +359,72 @@ class MyStack: Go： +使用两个队列实现 +```go +type MyStack struct { + //创建两个队列 + queue1 []int + queue2 []int +} + + +func Constructor() MyStack { + return MyStack{ //初始化 + queue1:make([]int,0), + queue2:make([]int,0), + } +} + + +func (this *MyStack) Push(x int) { + //先将数据存在queue2中 + this.queue2 = append(this.queue2,x) + //将queue1中所有元素移到queue2中，再将两个队列进行交换 + this.Move() +} + + +func (this *MyStack) Move(){ + if len(this.queue1) == 0{ + //交换，queue1置为queue2,queue2置为空 + this.queue1,this.queue2 = this.queue2,this.queue1 + }else{ + //queue1元素从头开始一个一个追加到queue2中 + this.queue2 = append(this.queue2,this.queue1[0]) + this.queue1 = this.queue1[1:] //去除第一个元素 + this.Move() //重复 + } +} + +func (this *MyStack) Pop() int { + val := this.queue1[0] + this.queue1 = this.queue1[1:] //去除第一个元素 + return val + +} + + +func (this *MyStack) Top() int { + return this.queue1[0] //直接返回 +} + + +func (this *MyStack) Empty() bool { +return len(this.queue1) == 0 +} + + +/** + * Your MyStack object will be instantiated and called as such: + * obj := Constructor(); + * obj.Push(x); + * param_2 := obj.Pop(); + * param_3 := obj.Top(); + * param_4 := obj.Empty(); + */ +``` + +使用一个队列实现 ```go type MyStack struct { queue []int//创建一个队列 From 20976616179c3960eb14dd3661b5dd8469cda481 Mon Sep 17 00:00:00 2001 From: Arthur Date: Sun, 31 Oct 2021 17:04:12 +0100 Subject: [PATCH 27/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=201002.=E6=9F=A5?= =?UTF-8?q?=E6=89=BE=E5=B8=B8=E7=94=A8=E5=AD=97=E7=AC=A6.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1002.查找常用字符.md | 52 +++++++++++++++++++++++++++++ 1 file changed, 52 insertions(+) diff --git a/problems/1002.查找常用字符.md b/problems/1002.查找常用字符.md index 44a02ceb..a8450f29 100644 --- a/problems/1002.查找常用字符.md +++ b/problems/1002.查找常用字符.md @@ -331,6 +331,58 @@ func commonChars(_ words: [String]) -> [String] { } ``` +C: +```c +void updateHashTable(int* hashTableOne, int* hashTableTwo) { + int i; + for(i = 0; i < 26; i++) { + hashTableOne[i] = hashTableOne[i] < hashTableTwo[i] ? hashTableOne[i] : hashTableTwo[i]; + } +} + +char ** commonChars(char ** words, int wordsSize, int* returnSize){ + //用来统计所有字母出现的最小频率 + int hashTable[26] = { 0 }; + //初始化返回的char**数组以及返回数组长度 + *returnSize = 0; + char** ret = (char**)malloc(sizeof(char*) * 100); + + //如果输入数组长度为0，则返回NULL + if(!wordsSize) + return NULL; + + int i; + //更新第一个单词的字母频率 + for(i = 0; i < strlen(words[0]); i++) + hashTable[words[0][i] - 'a']++; + //更新从第二个单词开始的字母频率 + for(i = 1; i < wordsSize; i++) { + //创建新的哈希表，记录新的单词的字母频率 + int newHashTable[26] = { 0 }; + int j; + for(j = 0; j < strlen(words[i]); j++) { + newHashTable[words[i][j] - 'a']++; + } + //更新原哈希表 + updateHashTable(hashTable, newHashTable); + } + + //将哈希表中的字符变为字符串放入ret中 + for(i = 0; i < 26; i++) { + if(hashTable[i]) { + int j; + for(j = 0; j < hashTable[i]; j++) { + char* tempString = (char*)malloc(sizeof(char) * 2); + tempString[0] = i + 'a'; + tempString[1] = '\0'; + ret[(*returnSize)++] = tempString; + } + } + } + return ret; +} +``` + ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) * B站视频：[代码随想录](https://space.bilibili.com/525438321) From d47af5291680d2b2ef0604e5780089c4c9a817b2 Mon Sep 17 00:00:00 2001 From: Arthur Date: Sun, 31 Oct 2021 17:09:17 +0100 Subject: [PATCH 28/35] =?UTF-8?q?=E6=9B=B4=E6=96=B0=201002.=E6=9F=A5?= =?UTF-8?q?=E6=89=BE=E5=B8=B8=E7=94=A8=E5=AD=97=E7=AC=A6.md=20C=E8=AF=AD?= =?UTF-8?q?=E8=A8=80=E7=89=88=E6=9C=AC=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1002.查找常用字符.md | 1 + 1 file changed, 1 insertion(+) diff --git a/problems/1002.查找常用字符.md b/problems/1002.查找常用字符.md index a8450f29..5a8d1093 100644 --- a/problems/1002.查找常用字符.md +++ b/problems/1002.查找常用字符.md @@ -333,6 +333,7 @@ func commonChars(_ words: [String]) -> [String] { C: ```c +//若两个哈希表定义为char数组（每个单词的最大长度不会超过100，因此可以用char表示），可以提高时间和空间效率 void updateHashTable(int* hashTableOne, int* hashTableTwo) { int i; for(i = 0; i < 26; i++) { From 60442907e8ebf10f507b39e0c4ac8f928bb3f8cc Mon Sep 17 00:00:00 2001 From: hailincai Date: Mon, 1 Nov 2021 07:41:14 -0400 Subject: [PATCH 29/35] =?UTF-8?q?=E4=BF=AE=E6=94=B9=E4=BA=86Java=20dp?= =?UTF-8?q?=E6=96=B9=E6=B3=95=E7=9A=84=E6=B3=A8=E9=87=8A=20bad=20data=20--?= =?UTF-8?q?>=20base=20data?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改了Java dp方法的注释 bad data --> base data --- problems/0122.买卖股票的最佳时机II.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) 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 41d83847..130bb51f 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 @@ -154,7 +154,7 @@ class Solution { // 动态规划 // [天数][是否持有股票] int[][] dp = new int[prices.length][2]; - // bad case + // base case dp[0][0] = 0; dp[0][1] = -prices[0]; From 22b77d6ce256830db00717ea8c58240fcec5a889 Mon Sep 17 00:00:00 2001 From: hailincai Date: Mon, 1 Nov 2021 07:51:12 -0400 Subject: [PATCH 30/35] =?UTF-8?q?0055=20=E4=BF=AE=E6=94=B9=E4=BB=A3?= =?UTF-8?q?=E7=A0=81=E5=88=9D=E5=A7=8B=E8=A6=86=E7=9B=96=E8=8C=83=E5=9B=B4?= =?UTF-8?q?=E7=9A=84=E5=80=BC=E5=92=8C=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 0055 修改代码初始覆盖范围的值和注释 --- problems/0055.跳跃游戏.md | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/problems/0055.跳跃游戏.md b/problems/0055.跳跃游戏.md index 816eb64b..29e09a64 100644 --- a/problems/0055.跳跃游戏.md +++ b/problems/0055.跳跃游戏.md @@ -92,8 +92,8 @@ class Solution { if (nums.length == 1) { return true; } - //覆盖范围 - int coverRange = nums[0]; + //覆盖范围, 初始覆盖范围应该是0，因为下面的迭代是从下标0开始的 + int coverRange = 0; //在覆盖范围内更新最大的覆盖范围 for (int i = 0; i <= coverRange; i++) { coverRange = Math.max(coverRange, i + nums[i]); From 250dd8b8f3dc0fbf5925588eac09ed352b5616cc Mon Sep 17 00:00:00 2001 From: a1045520 Date: Tue, 2 Nov 2021 15:35:16 +0800 Subject: [PATCH 31/35] =?UTF-8?q?=E5=A2=9E=E5=8A=A0=200203.=E7=A7=BB?= =?UTF-8?q?=E9=99=A4=E9=93=BE=E8=A1=A8=E5=85=83=E7=B4=A0=20Rust=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0203.移除链表元素.md | 35 +++++++++++++++++++++++++++++ 1 file changed, 35 insertions(+) diff --git a/problems/0203.移除链表元素.md b/problems/0203.移除链表元素.md index 6c52886a..188edc2e 100644 --- a/problems/0203.移除链表元素.md +++ b/problems/0203.移除链表元素.md @@ -358,6 +358,41 @@ func removeElements(head *ListNode, val int) *ListNode { } ``` +RUST: +```rust +// Definition for singly-linked list. +// #[derive(PartialEq, Eq, Clone, Debug)] +// pub struct ListNode { +// pub val: i32, +// pub next: Option> +// } +// +// impl ListNode { +// #[inline] +// fn new(val: i32) -> Self { +// ListNode { +// next: None, +// val +// } +// } +// } +impl Solution { + pub fn remove_elements(head: Option>, val: i32) -> Option> { + let mut head = head; + let mut dummy_head = ListNode::new(0); + let mut cur = &mut dummy_head; + while let Some(mut node) = head { + head = std::mem::replace(&mut node.next, None); + if node.val != val { + cur.next = Some(node); + cur = cur.next.as_mut().unwrap(); + } + } + dummy_head.next + } +} +``` + ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) * B站视频：[代码随想录](https://space.bilibili.com/525438321) From 0d5c3c34de875f845e324ed1772ec3e0db521bbd Mon Sep 17 00:00:00 2001 From: ArthurP Date: Wed, 3 Nov 2021 23:20:18 +0100 Subject: [PATCH 32/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200110.=E5=B9=B3?= =?UTF-8?q?=E8=A1=A1=E4=BA=8C=E5=8F=89=E6=A0=91.md=20C=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E9=80=92=E5=BD=92=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 30 ++++++++++++++++++++++++++++++ 1 file changed, 30 insertions(+) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index 43a49758..e22a12b3 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -619,6 +619,36 @@ var isBalanced = function(root) { }; ``` +##C +递归法： +```c +int getDepth(struct TreeNode* node) { + //如果结点不存在，返回0 + if(!node) + return 0; + //求出右子树深度 + int rightDepth = getDepth(node->right); + //求出左子树深度 + int leftDepth = getDepth(node->left); + //返回左右子树中的较大值+1 + return rightDepth > leftDepth ? rightDepth + 1 : leftDepth + 1; +} + +bool isBalanced(struct TreeNode* root) { + //递归结束条件为：传入结点为NULL，返回True + if(!root) + return 1; + //求出左右子树的深度 + int leftDepth = getDepth(root->left); + int rightDepth = getDepth(root->right); + int diff; + //若左右子树绝对值差距大于1，返回False + if((diff = leftDepth - rightDepth) > 1 || diff < -1) + return 0; + //检查左右子树是否为平衡二叉树 + return isBalanced(root->right) && isBalanced(root->left); +} +``` ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) From ad19a88ecf7559b2634f236c3a88a3ed15e37388 Mon Sep 17 00:00:00 2001 From: Arthur Date: Thu, 4 Nov 2021 10:21:03 +0000 Subject: [PATCH 33/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200110.=E5=B9=B3?= =?UTF-8?q?=E8=A1=A1=E4=BA=8C=E5=8F=89=E6=A0=91.md=20=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index 677fbc32..2e8c239f 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -635,7 +635,7 @@ var isBalanced = function(root) { }; ``` -##C +## C 递归法： ```c int getDepth(struct TreeNode* node) { From 18ba70884914ed10944da16d60bba1fda610262f Mon Sep 17 00:00:00 2001 From: Arthur Date: Thu, 4 Nov 2021 10:56:41 +0000 Subject: [PATCH 34/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200110.=E5=B9=B3?= =?UTF-8?q?=E8=A1=A1=E4=BA=8C=E5=8F=89=E6=A0=91.md=20C=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E8=BF=AD=E4=BB=A3=E6=B3=95?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 63 ++++++++++++++++++++++++++++++++ 1 file changed, 63 insertions(+) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index 2e8c239f..f3d70a5c 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -666,6 +666,69 @@ bool isBalanced(struct TreeNode* root) { } ``` +迭代法： +```c +//计算结点深度 +int getDepth(struct TreeNode* node) { + struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 10000); + int stackTop = 0; + if(node) + stack[stackTop++] = node; + int result = 0; + int depth = 0; + + while(stackTop) { + struct TreeNode* tempNode = stack[--stackTop]; + if(tempNode) { + depth++; + stack[stackTop++] = tempNode; + stack[stackTop++] = NULL; + if(tempNode->left) + stack[stackTop++] = tempNode->left; + if(tempNode->right) + stack[stackTop++] = tempNode->right; + result = result > depth ? result : depth; + } + else { + tempNode = stack[--stackTop]; + depth--; + } + } + + return result; +} + +bool isBalanced(struct TreeNode* root){ + //开辟栈空间 + struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 10000); + int stackTop = 0; + + //若根节点不存在，返回True + if(!root) + return 1; + + //将根节点入栈 + stack[stackTop++] = root; + //当栈中有元素时，进行遍历 + while(stackTop) { + //将栈顶元素出栈 + struct TreeNode* node = stack[--stackTop]; + //计算左右子树的深度 + int diff = getDepth(node->right) - getDepth(node->left); + //若深度的绝对值大于1，返回False + if(diff > 1 || diff < -1) + return 0; + //如果栈顶结点有左右结点，将左右结点入栈 + if(node->left) + stack[stackTop++] = node->left; + if(node->right) + stack[stackTop++] = node->right; + } + //若二叉树遍历结束后没有返回False，则返回True + return 1; +} +``` + ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) * B站视频：[代码随想录](https://space.bilibili.com/525438321) From a1004808796f6b97bd86753e1b9d859ebab4ff09 Mon Sep 17 00:00:00 2001 From: ArthurP Date: Sat, 6 Nov 2021 10:51:37 +0100 Subject: [PATCH 35/35] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200110.=E5=B9=B3?= =?UTF-8?q?=E8=A1=A1=E4=BA=8C=E5=8F=89=E6=A0=91.md=20C=E8=AF=AD=E8=A8=80?= =?UTF-8?q?=E8=BF=AD=E4=BB=A3=E6=B3=95=E6=B3=A8=E9=87=8A?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0110.平衡二叉树.md | 9 +++++++++ 1 file changed, 9 insertions(+) diff --git a/problems/0110.平衡二叉树.md b/problems/0110.平衡二叉树.md index f3d70a5c..b0ecf44a 100644 --- a/problems/0110.平衡二叉树.md +++ b/problems/0110.平衡二叉树.md @@ -670,26 +670,35 @@ bool isBalanced(struct TreeNode* root) { ```c //计算结点深度 int getDepth(struct TreeNode* node) { + //开辟栈空间 struct TreeNode** stack = (struct TreeNode**)malloc(sizeof(struct TreeNode*) * 10000); int stackTop = 0; + //若传入结点存在，将其入栈。若不存在，函数直接返回0 if(node) stack[stackTop++] = node; int result = 0; int depth = 0; + //当栈中有元素时，进行迭代遍历 while(stackTop) { + //取出栈顶元素 struct TreeNode* tempNode = stack[--stackTop]; + //若栈顶元素非NULL，则将深度+1 if(tempNode) { depth++; + //将栈顶元素再次入栈，添加NULL表示此结点已被遍历 stack[stackTop++] = tempNode; stack[stackTop++] = NULL; + //若栈顶元素有左右孩子，则将孩子结点入栈 if(tempNode->left) stack[stackTop++] = tempNode->left; if(tempNode->right) stack[stackTop++] = tempNode->right; + //更新结果 result = result > depth ? result : depth; } else { + //若为NULL，则代表当前结点已被遍历，深度-1 tempNode = stack[--stackTop]; depth--; }