From 31c9f09776b9ef6bf9e7861a32a43fde571fdcbf Mon Sep 17 00:00:00 2001 From: reoooh Date: Wed, 28 Jul 2021 23:57:26 +0800 Subject: [PATCH 01/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A00704=E4=BA=8C=E5=88=86?= =?UTF-8?q?=E6=9F=A5=E6=89=BE=20Ruby=20=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0704.二分查找.md | 42 ++++++++++++++++++++++++++++++++--- 1 file changed, 39 insertions(+), 3 deletions(-) diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md index cce471ae..66be52a0 100644 --- a/problems/0704.二分查找.md +++ b/problems/0704.二分查找.md @@ -235,8 +235,6 @@ class Solution: return -1 ``` - - **Go：** （版本一）左闭右闭区间 @@ -279,7 +277,7 @@ func search(nums []int, target int) int { } ``` -**javaScript:** +**JavaScript:** ```js @@ -316,7 +314,45 @@ var search = function(nums, target) { ``` +**Ruby:** +```ruby +# （版本一）左闭右闭区间 + +def search(nums, target) + left, right = 0, nums.length - 1 + while left <= right # 由于定义target在一个在左闭右闭的区间里，因此极限情况下存在left==right + middle = (left + right) / 2 + if nums[middle] > target + right = middle - 1 + elsif nums[middle] < target + left = middle + 1 + else + return middle # return兼具返回与跳出循环的作用 + end + end + -1 +end + +# （版本二）左闭右开区间 + +def search(nums, target) + left, right = 0, nums.length - 1 + while left < right # 由于定义target在一个在左闭右开的区间里，因此极限情况下right=left+1 + middle = (left + right) / 2 + if nums[middle] > target + right = middle + elsif nums[middle] < target + left = middle + 1 + else + return middle + end + end + -1 +end + +p search(nums,target) +``` From 4fcb27c7cf827d0c13bc261c4374cb5b199cbf1d Mon Sep 17 00:00:00 2001 From: reoooh Date: Thu, 29 Jul 2021 00:07:44 +0800 Subject: [PATCH 02/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A00704=E4=BA=8C=E5=88=86?= =?UTF-8?q?=E6=9F=A5=E6=89=BE=20Ruby=20=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0704.二分查找.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md index 66be52a0..caec38b1 100644 --- a/problems/0704.二分查找.md +++ b/problems/0704.二分查找.md @@ -337,7 +337,7 @@ end # （版本二）左闭右开区间 def search(nums, target) - left, right = 0, nums.length - 1 + left, right = 0, nums.length while left < right # 由于定义target在一个在左闭右开的区间里，因此极限情况下right=left+1 middle = (left + right) / 2 if nums[middle] > target From 2c7ea5a9f595acbebf03ba656d72ecc2b65d6a88 Mon Sep 17 00:00:00 2001 From: jiangjiang <366649270@qq.com> Date: Thu, 29 Jul 2021 00:46:31 +0800 Subject: [PATCH 03/53] =?UTF-8?q?Update=200053.=E6=9C=80=E5=A4=A7=E5=AD=90?= =?UTF-8?q?=E5=BA=8F=E5=92=8C=EF=BC=88=E5=8A=A8=E6=80=81=E8=A7=84=E5=88=92?= =?UTF-8?q?=EF=BC=89.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit add solution of golang for 0053.最大子序和 --- .../0053.最大子序和（动态规划）.md | 28 +++++++++++++++++++ 1 file changed, 28 insertions(+) diff --git a/problems/0053.最大子序和（动态规划）.md b/problems/0053.最大子序和（动态规划）.md index dd0e513b..ada378b2 100644 --- a/problems/0053.最大子序和（动态规划）.md +++ b/problems/0053.最大子序和（动态规划）.md @@ -138,7 +138,35 @@ class Solution: ``` Go： +```Go +// solution +// 1, dp +// 2, 贪心 +func maxSubArray(nums []int) int { + n := len(nums) + // 这里的dp[i] 表示，最大的连续子数组和，包含num[i] 元素 + dp := make([]int,n) + // 初始化，由于dp 状态转移方程依赖dp[0] + dp[0] = nums[0] + // 初始化最大的和 + mx := nums[0] + for i:=1;ib { + return a + } + return b +} +``` From 39f313ff0cec5c1ca94d56116a6b78f15882a56a Mon Sep 17 00:00:00 2001 From: wjjiang <48505670+Spongecaptain@users.noreply.github.com> Date: Thu, 29 Jul 2021 11:44:51 +0800 Subject: [PATCH 04/53] =?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 Go 中函数应该为小写 与 leetcode 保持一致。 --- problems/二叉树的递归遍历.md | 6 +++--- 1 file changed, 3 insertions(+), 3 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 2b5a44dd..220ccf65 100644 --- a/problems/二叉树的递归遍历.md +++ b/problems/二叉树的递归遍历.md @@ -222,7 +222,7 @@ Go： 前序遍历: ```go -func PreorderTraversal(root *TreeNode) (res []int) { +func preorderTraversal(root *TreeNode) (res []int) { var traversal func(node *TreeNode) traversal = func(node *TreeNode) { if node == nil { @@ -240,7 +240,7 @@ func PreorderTraversal(root *TreeNode) (res []int) { 中序遍历: ```go -func InorderTraversal(root *TreeNode) (res []int) { +func inorderTraversal(root *TreeNode) (res []int) { var traversal func(node *TreeNode) traversal = func(node *TreeNode) { if node == nil { @@ -257,7 +257,7 @@ func InorderTraversal(root *TreeNode) (res []int) { 后序遍历: ```go -func PostorderTraversal(root *TreeNode) (res []int) { +func postorderTraversal(root *TreeNode) (res []int) { var traversal func(node *TreeNode) traversal = func(node *TreeNode) { if node == nil { From 7029cf6b6eb760df09d39ce9a399f34016682ccf Mon Sep 17 00:00:00 2001 From: reoooh Date: Thu, 29 Jul 2021 13:49:02 +0800 Subject: [PATCH 05/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A00704=E4=BA=8C=E5=88=86?= =?UTF-8?q?=E6=9F=A5=E6=89=BE=20Ruby=20=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0704.二分查找.md | 2 -- 1 file changed, 2 deletions(-) diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md index caec38b1..ed0e2eed 100644 --- a/problems/0704.二分查找.md +++ b/problems/0704.二分查找.md @@ -350,8 +350,6 @@ def search(nums, target) end -1 end - -p search(nums,target) ``` From e9c5f17af52cbeee408e349d7a7655b5d711bc80 Mon Sep 17 00:00:00 2001 From: posper Date: Thu, 29 Jul 2021 18:48:56 +0800 Subject: [PATCH 06/53] =?UTF-8?q?724.=E5=AF=BB=E6=89=BE=E6=95=B0=E7=BB=84?= =?UTF-8?q?=E7=9A=84=E4=B8=AD=E5=BF=83=E4=B8=8B=E6=A0=87=20=E6=B7=BB?= =?UTF-8?q?=E5=8A=A0Java=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0724.寻找数组的中心索引.md | 19 ++++++++++++++++++- 1 file changed, 18 insertions(+), 1 deletion(-) diff --git a/problems/0724.寻找数组的中心索引.md b/problems/0724.寻找数组的中心索引.md index bf989979..09966e05 100644 --- a/problems/0724.寻找数组的中心索引.md +++ b/problems/0724.寻找数组的中心索引.md @@ -67,6 +67,24 @@ public: ## Java ```java +class Solution { + public int pivotIndex(int[] nums) { + int sum = 0; + for (int i = 0; i < nums.length; i++) { + sum += nums[i]; // 总和 + } + int leftSum = 0; + int rightSum = 0; + for (int i = 0; i < nums.length; i++) { + leftSum += nums[i]; + rightSum = sum - leftSum + nums[i]; // leftSum 里面已经有 nums[i]，多减了一次，所以加上 + if (leftSum == rightSum) { + return i; + } + } + return -1; // 不存在 + } +} ``` ## Python @@ -90,4 +108,3 @@ public: * 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
- From 3fd951ef5fec91b3dc9be7e57aa76daf83c35f87 Mon Sep 17 00:00:00 2001 From: posper Date: Thu, 29 Jul 2021 18:56:15 +0800 Subject: [PATCH 07/53] =?UTF-8?q?922.=20=E6=8C=89=E5=A5=87=E5=81=B6?= =?UTF-8?q?=E6=8E=92=E5=BA=8F=E6=95=B0=E7=BB=84II=20=E6=B7=BB=E5=8A=A0Java?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0922.按奇偶排序数组II.md | 26 +++++++++++++++++++++++- 1 file changed, 25 insertions(+), 1 deletion(-) diff --git a/problems/0922.按奇偶排序数组II.md b/problems/0922.按奇偶排序数组II.md index bb609c12..d4c1be7a 100644 --- a/problems/0922.按奇偶排序数组II.md +++ b/problems/0922.按奇偶排序数组II.md @@ -120,6 +120,31 @@ public: ## Java ```java +// 方法一 +class Solution { + public int[] sortArrayByParityII(int[] nums) { + // 分别存放 nums 中的奇数、偶数 + int len = nums.length; + int evenIndex = 0; + int oddIndex = 0; + int[] even = new int[len / 2]; + int[] odd = new int[len / 2]; + for (int i = 0; i < len; i++) { + if (nums[i] % 2 == 0) { + even[evenIndex++] = nums[i]; + } else { + odd[oddIndex++] = nums[i]; + } + } + // 把奇偶数组重新存回 nums + int index = 0; + for (int i = 0; i < even.length; i++) { + nums[index++] = even[i]; + nums[index++] = odd[i]; + } + return nums; + } +} ``` ## Python @@ -143,4 +168,3 @@ public: * 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
- From d8247fe8499f14497972fc32456f752a7bf6d574 Mon Sep 17 00:00:00 2001 From: posper Date: Thu, 29 Jul 2021 19:22:47 +0800 Subject: [PATCH 08/53] =?UTF-8?q?844.=E6=AF=94=E8=BE=83=E5=90=AB=E9=80=80?= =?UTF-8?q?=E6=A0=BC=E7=9A=84=E5=AD=97=E7=AC=A6=E4=B8=B2=20=E6=B7=BB?= =?UTF-8?q?=E5=8A=A0Java=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0844.比较含退格的字符串.md | 28 +++++++++++++++++++- 1 file changed, 27 insertions(+), 1 deletion(-) diff --git a/problems/0844.比较含退格的字符串.md b/problems/0844.比较含退格的字符串.md index 9f37959d..4b4ff63b 100644 --- a/problems/0844.比较含退格的字符串.md +++ b/problems/0844.比较含退格的字符串.md @@ -14,7 +14,7 @@ 给定 S 和 T 两个字符串，当它们分别被输入到空白的文本编辑器后，判断二者是否相等，并返回结果。 # 代表退格字符。 注意：如果对空文本输入退格字符，文本继续为空。 -  + 示例 1： * 输入：S = "ab#c", T = "ad#c" * 输出：true @@ -160,6 +160,32 @@ public: Java： +```java +// 普通方法（使用栈的思路） +class Solution { + public boolean backspaceCompare(String s, String t) { + StringBuilder ssb = new StringBuilder(); // 模拟栈 + StringBuilder tsb = new StringBuilder(); // 模拟栈 + // 分别处理两个 String + for (char c : s.toCharArray()) { + if (c != '#') { + ssb.append(c); // 模拟入栈 + } else if (ssb.length() > 0){ // 栈非空才能弹栈 + ssb.deleteCharAt(ssb.length() - 1); // 模拟弹栈 + } + } + for (char c : t.toCharArray()) { + if (c != '#') { + tsb.append(c); // 模拟入栈 + } else if (tsb.length() > 0){ // 栈非空才能弹栈 + tsb.deleteCharAt(tsb.length() - 1); // 模拟弹栈 + } + } + return ssb.toString().equals(tsb.toString()); + } +} +``` + Python： Go： From a6a8650b8a6524ed03cf599f1c32fc42c78cd8e5 Mon Sep 17 00:00:00 2001 From: posper Date: Thu, 29 Jul 2021 19:32:33 +0800 Subject: [PATCH 09/53] =?UTF-8?q?205.=E5=90=8C=E6=9E=84=E5=AD=97=E7=AC=A6?= =?UTF-8?q?=E4=B8=B2=20=E6=B7=BB=E5=8A=A0Java=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0205.同构字符串.md | 19 +++++++++++++++++++ 1 file changed, 19 insertions(+) diff --git a/problems/0205.同构字符串.md b/problems/0205.同构字符串.md index 4e963ece..06ea5701 100644 --- a/problems/0205.同构字符串.md +++ b/problems/0205.同构字符串.md @@ -68,6 +68,25 @@ public: ## Java ```java +class Solution { + public boolean isIsomorphic(String s, String t) { + Map map1 = new HashMap<>(); + Map map2 = new HashMap<>(); + for (int i = 0, j = 0; i < s.length(); i++, j++) { + if (!map1.containsKey(s.charAt(i))) { + map1.put(s.charAt(i), t.charAt(j)); // map1保存 s[i] 到 t[j]的映射 + } + if (!map2.containsKey(t.charAt(j))) { + map2.put(t.charAt(j), s.charAt(i)); // map1保存 t[j] 到 s[i]的映射 + } + // 无法映射，返回 false + if (map1.get(s.charAt(i)) != t.charAt(j) || map2.get(t.charAt(j)) != s.charAt(i)) { + return false; + } + } + return true; + } +} ``` ## Python From 9e08c0e5de17773b345f0cc59fb661b333bd946f Mon Sep 17 00:00:00 2001 From: posper Date: Thu, 29 Jul 2021 19:36:37 +0800 Subject: [PATCH 10/53] =?UTF-8?q?205.=E5=90=8C=E6=9E=84=E5=AD=97=E7=AC=A6?= =?UTF-8?q?=E4=B8=B2=20=E6=B7=BB=E5=8A=A0Java=E7=89=88=E6=9C=AC=EF=BC=88?= =?UTF-8?q?=E6=B3=A8=E9=87=8A=E4=BF=AE=E6=94=B9=EF=BC=89?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0205.同构字符串.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0205.同构字符串.md b/problems/0205.同构字符串.md index 06ea5701..7ffa3fbf 100644 --- a/problems/0205.同构字符串.md +++ b/problems/0205.同构字符串.md @@ -77,7 +77,7 @@ class Solution { map1.put(s.charAt(i), t.charAt(j)); // map1保存 s[i] 到 t[j]的映射 } if (!map2.containsKey(t.charAt(j))) { - map2.put(t.charAt(j), s.charAt(i)); // map1保存 t[j] 到 s[i]的映射 + map2.put(t.charAt(j), s.charAt(i)); // map2保存 t[j] 到 s[i]的映射 } // 无法映射，返回 false if (map1.get(s.charAt(i)) != t.charAt(j) || map2.get(t.charAt(j)) != s.charAt(i)) { From 1521e4016537243bd2a6d5898f2099e53f588d42 Mon Sep 17 00:00:00 2001 From: posper Date: Thu, 29 Jul 2021 20:00:16 +0800 Subject: [PATCH 11/53] =?UTF-8?q?31.=E4=B8=8B=E4=B8=80=E4=B8=AA=E6=8E=92?= =?UTF-8?q?=E5=88=97=20=E6=B7=BB=E5=8A=A0Java=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0031.下一个排列.md | 18 ++++++++++++++++++ 1 file changed, 18 insertions(+) diff --git a/problems/0031.下一个排列.md b/problems/0031.下一个排列.md index fbe31eb3..53e6644d 100644 --- a/problems/0031.下一个排列.md +++ b/problems/0031.下一个排列.md @@ -99,6 +99,24 @@ public: ## Java ```java +class Solution { + public void nextPermutation(int[] nums) { + for (int i = nums.length - 1; i >= 0; i--) { + for (int j = nums.length - 1; j > i; j--) { + if (nums[j] > nums[i]) { + // 交换 + int temp = nums[i]; + nums[i] = nums[j]; + nums[j] = temp; + // [i + 1, nums.length) 内元素升序排序 + Arrays.sort(nums, i + 1, nums.length); + return; + } + } + } + Arrays.sort(nums); // 不存在下一个更大的排列，则将数字重新排列成最小的排列（即升序排列）。 + } +} ``` ## Python From 0b2312ffec67c90a53feb938a6df5444c41fbc1a Mon Sep 17 00:00:00 2001 From: Kelvin Date: Thu, 29 Jul 2021 09:02:15 -0400 Subject: [PATCH 12/53] =?UTF-8?q?Update=200404.=E5=B7=A6=E5=8F=B6=E5=AD=90?= =?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 添加python3 迭代方法的代码. Q: 我为什么要更改当前已有的递归代码? A: 因为我发现当前版本的递归代码跟Carl哥的逻辑不是特别一样. 我承认解题思路千变万化, 但是个人感觉最好最好还是尽量跟Carl的思路一样, 这样方便初学者阅读不同语言编写出来的代码. --- problems/0404.左叶子之和.md | 56 +++++++++++++++++++++++--------- 1 file changed, 41 insertions(+), 15 deletions(-) diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md index aa758367..2a76a461 100644 --- a/problems/0404.左叶子之和.md +++ b/problems/0404.左叶子之和.md @@ -205,25 +205,51 @@ class Solution { Python： -```Python + **递归** -# 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 +```python class Solution: def sumOfLeftLeaves(self, root: TreeNode) -> int: - self.res=0 - def areleftleaves(root): - if not root:return - if root.left and (not root.left.left) and (not root.left.right):self.res+=root.left.val - areleftleaves(root.left) - areleftleaves(root.right) - areleftleaves(root) - return self.res + 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 # 中 + + return cur_left_leaf_val + left_left_leaves_sum + right_left_leaves_sum ``` + +**迭代** +```python +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 +``` + Go： > 递归法 From 2f412e5c7d3b41b66946d92e699330cb7e5e5036 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Thu, 29 Jul 2021 11:18:26 -0400 Subject: [PATCH 13/53] =?UTF-8?q?=E6=9B=B4=E6=96=B0=20513.=E6=89=BE?= =?UTF-8?q?=E6=A0=91=E5=B7=A6=E4=B8=8B=E8=A7=92=E7=9A=84=E5=80=BC=20python?= =?UTF-8?q?=20=E9=83=A8=E5=88=86=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 1. 更改python 递归代码: 上一版本python代码跟卡哥思路不一样, 感觉把不同的解题方法放在这里不方便初学者学习. 所以建议更改. 2. 添加迭代方法代码. --- problems/0513.找树左下角的值.md | 61 +++++++++++++++++++++----- 1 file changed, 49 insertions(+), 12 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 17d15fde..391118ab 100644 --- a/problems/0513.找树左下角的值.md +++ b/problems/0513.找树左下角的值.md @@ -274,8 +274,9 @@ class Solution { Python： + +**递归法 - 回溯** ```python -//递归法 # Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): @@ -284,17 +285,53 @@ Python： # self.right = right class Solution: def findBottomLeftValue(self, root: TreeNode) -> int: - depth=0 - self.res=[] - def level(root,depth): - if not root:return - if depth==len(self.res): - self.res.append([]) - self.res[depth].append(root.val) - level(root.left,depth+1) - level(root.right,depth+1) - level(root,depth) - return self.res[-1][0] + max_depth = -float("INF") + max_left_value = -float("INF") + + def __traversal(root, left_len): + nonlocal max_depth, max_left_value + + if not root.left and not root.right: + if left_len > max_depth: + max_depth = left_len + max_left_value = root.val + return + + if root.left: + left_len += 1 + __traversal(root.left, left_len) + left_len -= 1 + + if root.right: + left_len += 1 + __traversal(root.right, left_len) + left_len -= 1 + return + + __traversal(root, 0) + + return max_left_value +``` + +**迭代法 - 层序遍历** +```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 ``` Go： From 8c53c48ea87fc2f02934799da6f659a7647eb268 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Thu, 29 Jul 2021 11:21:14 -0400 Subject: [PATCH 14/53] =?UTF-8?q?Revert=20"=E6=9B=B4=E6=96=B0=20513.?= =?UTF-8?q?=E6=89=BE=E6=A0=91=E5=B7=A6=E4=B8=8B=E8=A7=92=E7=9A=84=E5=80=BC?= =?UTF-8?q?=20python=20=E9=83=A8=E5=88=86=E4=BB=A3=E7=A0=81"?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit This reverts commit 2f412e5c7d3b41b66946d92e699330cb7e5e5036. --- problems/0513.找树左下角的值.md | 61 +++++--------------------- 1 file changed, 12 insertions(+), 49 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 391118ab..17d15fde 100644 --- a/problems/0513.找树左下角的值.md +++ b/problems/0513.找树左下角的值.md @@ -274,9 +274,8 @@ class Solution { Python： - -**递归法 - 回溯** ```python +//递归法 # Definition for a binary tree node. # class TreeNode: # def __init__(self, val=0, left=None, right=None): @@ -285,53 +284,17 @@ Python： # self.right = right class Solution: def findBottomLeftValue(self, root: TreeNode) -> int: - max_depth = -float("INF") - max_left_value = -float("INF") - - def __traversal(root, left_len): - nonlocal max_depth, max_left_value - - if not root.left and not root.right: - if left_len > max_depth: - max_depth = left_len - max_left_value = root.val - return - - if root.left: - left_len += 1 - __traversal(root.left, left_len) - left_len -= 1 - - if root.right: - left_len += 1 - __traversal(root.right, left_len) - left_len -= 1 - return - - __traversal(root, 0) - - return max_left_value -``` - -**迭代法 - 层序遍历** -```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 + depth=0 + self.res=[] + def level(root,depth): + if not root:return + if depth==len(self.res): + self.res.append([]) + self.res[depth].append(root.val) + level(root.left,depth+1) + level(root.right,depth+1) + level(root,depth) + return self.res[-1][0] ``` Go： From 49d55fb1c8c26f668c1676075ac10a21b527ab92 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Thu, 29 Jul 2021 11:25:33 -0400 Subject: [PATCH 15/53] =?UTF-8?q?=E6=9B=B4=E6=96=B0=20513.=E6=89=BE?= =?UTF-8?q?=E6=A0=91=E5=B7=A6=E4=B8=8B=E8=A7=92=E7=9A=84=E5=80=BC=20python?= =?UTF-8?q?=E9=83=A8=E5=88=86=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 1. 上一版本 递归python代码跟题解的方式不同. 个人认为不同方法做出的代码没有必要放在题解里, 建议修改成我提供的python版本. 2. 增加Python3 迭代部分的代码. --- problems/0513.找树左下角的值.md | 66 +++++++++++++++++++------- 1 file changed, 48 insertions(+), 18 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 17d15fde..e83fcc18 100644 --- a/problems/0513.找树左下角的值.md +++ b/problems/0513.找树左下角的值.md @@ -274,27 +274,57 @@ class Solution { Python： + +**递归 - 回溯** ```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 findBottomLeftValue(self, root: TreeNode) -> int: - depth=0 - self.res=[] - def level(root,depth): - if not root:return - if depth==len(self.res): - self.res.append([]) - self.res[depth].append(root.val) - level(root.left,depth+1) - level(root.right,depth+1) - level(root,depth) - return self.res[-1][0] + max_depth = -float("INF") + max_left_value = -float("INF") + + def __traversal(root, left_len): + nonlocal max_depth, max_left_value + + if not root.left and not root.right: + if left_len > max_depth: + max_depth = left_len + max_left_value = root.val + return + + if root.left: + left_len += 1 + __traversal(root.left, left_len) + left_len -= 1 + + if root.right: + left_len += 1 + __traversal(root.right, left_len) + left_len -= 1 + return + + __traversal(root, 0) + + return max_left_value +``` +**迭代 - 层序遍历** +```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 ``` Go： From 1dd4a529fcc7e7d9fb2de170fbd628d9f2af6f7f Mon Sep 17 00:00:00 2001 From: Kelvin Date: Thu, 29 Jul 2021 13:50:56 -0400 Subject: [PATCH 16/53] =?UTF-8?q?Update=200513.=E6=89=BE=E6=A0=91=E5=B7=A6?= =?UTF-8?q?=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 更改变量名称 --- problems/0513.找树左下角的值.md | 34 +++++++++++--------------- 1 file changed, 14 insertions(+), 20 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 e83fcc18..27c6e83c 100644 --- a/problems/0513.找树左下角的值.md +++ b/problems/0513.找树左下角的值.md @@ -280,31 +280,25 @@ Python： class Solution: def findBottomLeftValue(self, root: TreeNode) -> int: max_depth = -float("INF") - max_left_value = -float("INF") - - def __traversal(root, left_len): - nonlocal max_depth, max_left_value + leftmost_val = 0 + def __traverse(root, cur_depth): + nonlocal max_depth, leftmost_val if not root.left and not root.right: - if left_len > max_depth: - max_depth = left_len - max_left_value = root.val - return - + if cur_depth > max_depth: + max_depth = cur_depth + leftmost_val = root.val if root.left: - left_len += 1 - __traversal(root.left, left_len) - left_len -= 1 - + cur_depth += 1 + __traverse(root.left, cur_depth) + cur_depth -= 1 if root.right: - left_len += 1 - __traversal(root.right, left_len) - left_len -= 1 - return - - __traversal(root, 0) + cur_depth += 1 + __traverse(root.right, cur_depth) + cur_depth -= 1 - return max_left_value + __traverse(root, 0) + return leftmost_val ``` **迭代 - 层序遍历** ```python From 440e04fffe7bc5cfacd11653fc6f0d3704382767 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Thu, 29 Jul 2021 15:50:46 -0400 Subject: [PATCH 17/53] =?UTF-8?q?=E6=9B=B4=E6=94=B9=200112.=E8=B7=AF?= =?UTF-8?q?=E5=BE=84=E6=80=BB=E5=92=8C.md=20Python3=20=E4=BB=A3=E7=A0=81.?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 0112.路径总和: 1. 修改了 python3 注释, 更改代码格式. 2. 增加了迭代方法Python3代码. 0113.路径总和-ii: 上一版python3代码的问题: (1)注释一般不用// (2)代码格式有些不规范. (3)内层函数命名也不是特别的informative. 新版本代码修改了上述问题. 本次提交代码均已通过leetcode测试. --- problems/0112.路径总和.md | 113 ++++++++++++++++++++++------------ 1 file changed, 72 insertions(+), 41 deletions(-) diff --git a/problems/0112.路径总和.md b/problems/0112.路径总和.md index ae9a9267..474e17da 100644 --- a/problems/0112.路径总和.md +++ b/problems/0112.路径总和.md @@ -416,6 +416,8 @@ class Solution { Python： 0112.路径总和 + +**递归** ```python # Definition for a binary tree node. # class TreeNode: @@ -424,28 +426,56 @@ Python： # self.left = left # self.right = right -// 递归法 - 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 + 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 + 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 + 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) + + if root == None: + return False # 别忘记处理空TreeNode + else: + return isornot(root, targetSum - root.val) ``` +**迭代 - 层序遍历** +```python +class Solution: + def hasPathSum(self, root: TreeNode, targetSum: 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 +``` 0113.路径总和-ii + +**递归** ```python # Definition for a binary tree node. # class TreeNode: @@ -453,35 +483,36 @@ class Solution: # self.val = val # self.left = left # self.right = right -//递归法 class Solution: def pathSum(self, root: TreeNode, targetSum: int) -> List[List[int]]: - path=[] - res=[] - def pathes(root,targetSum): - if (not root.left) and (not root.right) and targetSum == 0: // 遇到叶子节点，并且计数为0 - res.append(path[:]) //找到一种路径，记录到res中，注意必须是path[:]而不是path - return - if (not root.left) and (not root.right):return // 遇到叶子节点直接返回 - if root.left: //左 - targetSum -= root.left.val - path.append(root.left.val) //递归前记录节点 - pathes(root.left,targetSum) //递归 - targetSum += root.left.val //回溯 - path.pop() //回溯 - if root.right: //右 - targetSum -= root.right.val - path.append(root.right.val) //递归前记录节点 - pathes(root.right,targetSum) //递归 - targetSum += root.right.val //回溯 - path.pop() //回溯 - return - - if root == None:return [] //处理空TreeNode - else: - path.append(root.val) //首先处理根节点 - pathes(root,targetSum-root.val) - return res + + def traversal(cur_node, remain): + if not cur_node.left and not cur_node.right and remain == 0: + result.append(path[:]) + return + + if not cur_node.left and not cur_node.right: return + + if cur_node.left: + path.append(cur_node.left.val) + remain -= cur_node.left.val + traversal(cur_node.left, remain) + path.pop() + remain += cur_node.left.val + + if cur_node.right: + path.append(cur_node.right.val) + remain -= cur_node.right.val + traversal(cur_node.right, remain) + path.pop() + remain += cur_node.right.val + + result, path = [], [] + if not root: + return [] + path.append(root.val) + traversal(root, targetSum - root.val) + return result ``` Go： From d21853a3b22763c4d1cbe02ab6a2baba349348cb Mon Sep 17 00:00:00 2001 From: Jack <965555169@qq.com> Date: Fri, 30 Jul 2021 11:07:38 +0800 Subject: [PATCH 18/53] =?UTF-8?q?=E6=9B=B4=E6=96=B0718.=20=E6=9C=80?= =?UTF-8?q?=E9=95=BF=E9=87=8D=E5=A4=8D=E5=AD=90=E6=95=B0=E7=BB=84=20JavaSc?= =?UTF-8?q?ript=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0718.最长重复子数组.md | 27 ++++++++++++++++++++++++++ 1 file changed, 27 insertions(+) diff --git a/problems/0718.最长重复子数组.md b/problems/0718.最长重复子数组.md index 9ee94a3d..959ed39e 100644 --- a/problems/0718.最长重复子数组.md +++ b/problems/0718.最长重复子数组.md @@ -252,6 +252,33 @@ func findLength(A []int, B []int) int { } ``` +JavaScript： + +> 动态规划 + +```javascript +const findLength = (A, B) => { + // A、B数组的长度 + const [m, n] = [A.length, B.length]; + // dp数组初始化，都初始化为0 + const dp = new Array(m + 1).fill(0).map(x => new Array(n + 1).fill(0)); + // 初始化最大长度为0 + let res = 0; + for (let i = 1; i <= m; i++) { + for (let j = 1; j <= n; j++) { + // 遇到A[i - 1] === B[j - 1]，则更新dp数组 + if (A[i - 1] === B[j - 1]) { + dp[i][j] = dp[i - 1][j - 1] + 1; + } + // 更新res + res = dp[i][j] > res ? dp[i][j] : res; + } + } + // 遍历完成，返回res + return res; +}; +``` + ----------------------- From 2119e411b73d818baaec8ca8140c7b619358bd54 Mon Sep 17 00:00:00 2001 From: xsduan98 Date: Fri, 30 Jul 2021 18:21:18 +0800 Subject: [PATCH 19/53] =?UTF-8?q?657.=20=E6=9C=BA=E5=99=A8=E4=BA=BA?= =?UTF-8?q?=E8=83=BD=E5=90=A6=E8=BF=94=E5=9B=9E=E5=8E=9F=E7=82=B9=20?= =?UTF-8?q?=E6=B7=BB=E5=8A=A0Java=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0657.机器人能否返回原点.md | 15 +++++++++++++++ 1 file changed, 15 insertions(+) diff --git a/problems/0657.机器人能否返回原点.md b/problems/0657.机器人能否返回原点.md index cd26836c..b1fb6f21 100644 --- a/problems/0657.机器人能否返回原点.md +++ b/problems/0657.机器人能否返回原点.md @@ -71,6 +71,21 @@ public: ## Java ```java +// 时间复杂度：O(n) +// 空间复杂度：如果采用 toCharArray，则是 O（n）;如果使用 charAt，则是 O(1) +class Solution { + public boolean judgeCircle(String moves) { + int x = 0; + int y = 0; + for (char c : moves.toCharArray()) { + if (c == 'U') y++; + if (c == 'D') y--; + if (c == 'L') x++; + if (c == 'R') x--; + } + return x == 0 && y == 0; + } +} ``` ## Python From 58ed197ba94f17646969ddf0e45ed851f5b03961 Mon Sep 17 00:00:00 2001 From: xsduan98 Date: Fri, 30 Jul 2021 18:54:43 +0800 Subject: [PATCH 20/53] =?UTF-8?q?1365.=E6=9C=89=E5=A4=9A=E5=B0=91=E5=B0=8F?= =?UTF-8?q?=E4=BA=8E=E5=BD=93=E5=89=8D=E6=95=B0=E5=AD=97=E7=9A=84=E6=95=B0?= =?UTF-8?q?=E5=AD=97=20=E6=B7=BB=E5=8A=A0Java=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...据数字二进制下1的数目排序.md | 48 +++++++++++++++++-- 1 file changed, 45 insertions(+), 3 deletions(-) diff --git a/problems/1356.根据数字二进制下1的数目排序.md b/problems/1356.根据数字二进制下1的数目排序.md index e464f716..a3d724fa 100644 --- a/problems/1356.根据数字二进制下1的数目排序.md +++ b/problems/1356.根据数字二进制下1的数目排序.md @@ -16,7 +16,7 @@ 换而言之，对于每个 nums[i] 你必须计算出有效的 j 的数量，其中 j 满足 j != i 且 nums[j] < nums[i] 。 以数组形式返回答案。 -  + 示例 1： 输入：nums = [8,1,2,2,3] @@ -35,7 +35,7 @@ 示例 3： 输入：nums = [7,7,7,7] 输出：[0,0,0,0] -  + 提示： * 2 <= nums.length <= 500 * 0 <= nums[i] <= 100 @@ -120,8 +120,51 @@ public: ## Java ```java +/** +* 解法一：暴力 +* 时间复杂度：O(n^2) +* 空间复杂度：O(n) +*/ +class Solution { + public int[] smallerNumbersThanCurrent(int[] nums) { + int[] res = new int[nums.length]; + for (int i = 0; i < nums.length; i++) { + for (int j = 0; j < nums.length; j++) { + if (nums[j] < nums[i] && j != i) { // 注意 j 不能和 i 重合 + res[i]++; + } + } + } + return res; + } +} ``` +```java +/** +* 优化：排序 + 哈希表 +* 时间复杂度：O(nlogn) +* 空间复杂度：O(n) +*/ +class Solution { + public int[] smallerNumbersThanCurrent(int[] nums) { + int[] res = Arrays.copyOf(nums, nums.length); + Arrays.sort(res); // 是对 res 排序，nums 中顺序还要保持 + int[] hash = new int[101]; // 使用哈希表，记录比当前元素小的元素个数 + for (int i = res.length - 1; i >= 0; i--) { // 注意：从后向前 + hash[res[i]] = i; // 排序后，当前下标即表示比当前元素小的元素个数 + } + // 此时 hash中保存的每一个元素数值 便是 小于这个数值的个数 + for (int i = 0; i < res.length; i++) { + res[i] = hash[nums[i]]; + } + return res; + } +} +``` + + + ## Python ```python @@ -143,4 +186,3 @@ public: * 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ)
- From 220ceea1edd9f5e4299eb5e6814febecb9a60573 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Fri, 30 Jul 2021 15:14:16 -0400 Subject: [PATCH 21/53] =?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 上一版本隐藏了些细节. 代码格式不遵循PEP8 新版本代码改进格式并且加上注释. 同时每一步都是按照题解写出, 虽然比起上版本方法略显冗长, 但是更加利于理解. 方便初学者学习. --- ...序与后序遍历序列构造二叉树.md | 77 ++++++++++++------- 1 file changed, 51 insertions(+), 26 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 4c5a70a0..4bdc228d 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 @@ -654,43 +654,68 @@ class Solution { ``` Python： + 105.从前序与中序遍历序列构造二叉树 ```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 -//递归法 class Solution: def buildTree(self, preorder: List[int], inorder: List[int]) -> TreeNode: - if not preorder: return None //特殊情况 - root = TreeNode(preorder[0]) //新建父节点 - p=inorder.index(preorder[0]) //找到父节点在中序遍历的位置(因为没有重复的元素，才可以这样找) - root.left = self.buildTree(preorder[1:p+1],inorder[:p]) //注意左节点时分割中序数组和前续数组的开闭环 - root.right = self.buildTree(preorder[p+1:],inorder[p+1:]) //分割中序数组和前续数组 - return root + # 第一步: 特殊情况讨论: 树为空. 或者说是递归终止条件 + if not preorder: + return None + + # 第二步: 前序遍历的第一个就是当前的中间节点. + root_val = preorder[0] + root = TreeNode(root_val) + + # 第三步: 找切割点. + separator_idx = inorder.index(root_val) + + # 第四步: 切割inorder数组. 得到inorder数组的左,右半边. + inorder_left = inorder[:separator_idx] + inorder_right = inorder[separator_idx + 1:] + + # 第五步: 切割preorder数组. 得到preorder数组的左,右半边. + # ⭐️ 重点1: 中序数组大小一定跟前序数组大小是相同的. + preorder_left = preorder[1:1 + len(inorder_left)] + preorder_right = preorder[1 + len(inorder_left):] + + # 第六步: 递归 + root.left = self.buildTree(preorder_left, inorder_left) + root.right = self.buildTree(preorder_right, inorder_right) + + return root ``` 106.从中序与后序遍历序列构造二叉树 ```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 -//递归法 class Solution: def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode: - if not postorder: return None //特殊情况 - root = TreeNode(postorder[-1]) //新建父节点 - p=inorder.index(postorder[-1]) //找到父节点在中序遍历的位置*因为没有重复的元素，才可以这样找 - root.left = self.buildTree(inorder[:p],postorder[:p]) //分割中序数组和后续数组 - root.right = self.buildTree(inorder[p+1:],postorder[p:-1]) //注意右节点时分割中序数组和后续数组的开闭环 - return root + # 第一步: 特殊情况讨论: 树为空. (递归终止条件) + if not postorder: + return None + + # 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. + separator_idx = inorder.index(root_val) + + # 第四步: 切割inorder数组. 得到inorder数组的左,右半边. + inorder_left = inorder[:separator_idx] + inorder_right = inorder[separator_idx + 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. + postorder_left = postorder[:len(inorder_left)] + postorder_right = postorder[len(inorder_left): len(postorder) - 1] + + # 第六步: 递归 + root.left = self.buildTree(inorder_left, postorder_left) + root.right = self.buildTree(inorder_right, postorder_right) + + return root ``` Go： > 106 从中序与后序遍历序列构造二叉树 From 4764d993b26a3a912646f810095c49142d497be8 Mon Sep 17 00:00:00 2001 From: posper Date: Sat, 31 Jul 2021 10:20:56 +0800 Subject: [PATCH 22/53] =?UTF-8?q?657.=E6=9C=BA=E5=99=A8=E4=BA=BA=E8=83=BD?= =?UTF-8?q?=E5=90=A6=E8=BF=94=E5=9B=9E=E5=8E=9F=E7=82=B9=20=E6=B7=BB?= =?UTF-8?q?=E5=8A=A0python3=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0657.机器人能否返回原点.md | 14 ++++++++++++++ 1 file changed, 14 insertions(+) diff --git a/problems/0657.机器人能否返回原点.md b/problems/0657.机器人能否返回原点.md index b1fb6f21..ba682a33 100644 --- a/problems/0657.机器人能否返回原点.md +++ b/problems/0657.机器人能否返回原点.md @@ -91,6 +91,20 @@ class Solution { ## Python ```python +class Solution: + def judgeCircle(self, moves: str) -> bool: + x = 0 # 记录当前位置 + y = 0 + for i in range(len(moves)): + if (moves[i] == 'U'): + y += 1 + if (moves[i] == 'D'): + y -= 1 + if (moves[i] == 'L'): + x += 1 + if (moves[i] == 'R'): + x -= 1 + return x == 0 and y == 0 ``` ## Go From 0e832d518357bb239d84e294e08a3ccecec669ea Mon Sep 17 00:00:00 2001 From: posper Date: Sat, 31 Jul 2021 10:28:14 +0800 Subject: [PATCH 23/53] =?UTF-8?q?657.=E6=9C=BA=E5=99=A8=E4=BA=BA=E8=83=BD?= =?UTF-8?q?=E5=90=A6=E8=BF=94=E5=9B=9E=E5=8E=9F=E7=82=B9=20=E6=B7=BB?= =?UTF-8?q?=E5=8A=A0JavaScript=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0657.机器人能否返回原点.md | 15 +++++++++++++++ 1 file changed, 15 insertions(+) diff --git a/problems/0657.机器人能否返回原点.md b/problems/0657.机器人能否返回原点.md index ba682a33..805a81e6 100644 --- a/problems/0657.机器人能否返回原点.md +++ b/problems/0657.机器人能否返回原点.md @@ -91,6 +91,8 @@ class Solution { ## Python ```python +# 时间复杂度：O(n) +# 空间复杂度：O(1) class Solution: def judgeCircle(self, moves: str) -> bool: x = 0 # 记录当前位置 @@ -115,6 +117,19 @@ class Solution: ## JavaScript ```js +// 时间复杂度：O(n) +// 空间复杂度：O(1) +var judgeCircle = function(moves) { + var x = 0; // 记录当前位置 + var y = 0; + for (var i = 0; i < moves.length; i++) { + if (moves[i] == 'U') y++; + if (moves[i] == 'D') y--; + if (moves[i] == 'L') x++; + if (moves[i] == 'R') x--; + } + return x == 0 && y == 0; +}; ``` ----------------------- From 4851af73ab4ed58b71c1c6d1223c39d2483816ee Mon Sep 17 00:00:00 2001 From: posper Date: Sat, 31 Jul 2021 10:36:59 +0800 Subject: [PATCH 24/53] =?UTF-8?q?657.=E6=9C=BA=E5=99=A8=E4=BA=BA=E8=83=BD?= =?UTF-8?q?=E5=90=A6=E8=BF=94=E5=9B=9E=E5=8E=9F=E7=82=B9=20=E6=B7=BB?= =?UTF-8?q?=E5=8A=A0go=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0657.机器人能否返回原点.md | 19 +++++++++++++++++++ 1 file changed, 19 insertions(+) diff --git a/problems/0657.机器人能否返回原点.md b/problems/0657.机器人能否返回原点.md index 805a81e6..9b43ea6c 100644 --- a/problems/0657.机器人能否返回原点.md +++ b/problems/0657.机器人能否返回原点.md @@ -112,6 +112,25 @@ class Solution: ## Go ```go +func judgeCircle(moves string) bool { + x := 0 + y := 0 + for i := 0; i < len(moves); i++ { + if moves[i] == 'U' { + y++ + } + if moves[i] == 'D' { + y-- + } + if moves[i] == 'L' { + x++ + } + if moves[i] == 'R' { + x-- + } + } + return x == 0 && y == 0; +} ``` ## JavaScript From 4f5ebdabca741e124269cb5d139eaf816b5e8f4a Mon Sep 17 00:00:00 2001 From: reoooh Date: Sun, 1 Aug 2021 00:01:37 +0800 Subject: [PATCH 25/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A00206=E7=BF=BB=E8=BD=AC?= =?UTF-8?q?=E9=93=BE=E8=A1=A8=20Ruby=20=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0206.翻转链表.md | 43 +++++++++++++++++++++++++++++++++++ 1 file changed, 43 insertions(+) diff --git a/problems/0206.翻转链表.md b/problems/0206.翻转链表.md index 963d7916..4efdda3b 100644 --- a/problems/0206.翻转链表.md +++ b/problems/0206.翻转链表.md @@ -275,7 +275,50 @@ var reverseList = function(head) { }; ``` +Ruby: +```ruby +# 双指针 +# Definition for singly-linked list. +# class ListNode +# attr_accessor :val, :next +# def initialize(val = 0, _next = nil) +# @val = val +# @next = _next +# end +# end +def reverse_list(head) + # return nil if head.nil? # 循环判断条件亦能起到相同作用因此不必单独判断 + cur, per = head, nil + until cur.nil? + tem = cur.next + cur.next = per + per = cur + cur = tem + end + per +end + +# 递归 +# Definition for singly-linked list. +# class ListNode +# attr_accessor :val, :next +# def initialize(val = 0, _next = nil) +# @val = val +# @next = _next +# end +# end +def reverse_list(head) + reverse(nil, head) +end + +def reverse(pre, cur) + return pre if cur.nil? + tem = cur.next + cur.next = pre + reverse(cur, tem) # 通过递归实现双指针法中的更新操作 +end +``` ----------------------- From 932d27754cb34420926a41176f6fedb860768b44 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Sat, 31 Jul 2021 23:04:46 -0400 Subject: [PATCH 26/53] =?UTF-8?q?Update=200617.=E5=90=88=E5=B9=B6=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 修改逻辑与题解相同. --- problems/0617.合并二叉树.md | 25 ++++++++++++++++++------- 1 file changed, 18 insertions(+), 7 deletions(-) diff --git a/problems/0617.合并二叉树.md b/problems/0617.合并二叉树.md index 19b58bd3..c5eb7594 100644 --- a/problems/0617.合并二叉树.md +++ b/problems/0617.合并二叉树.md @@ -312,6 +312,8 @@ class Solution { ``` Python： + +**递归法 - 前序遍历** ```python # Definition for a binary tree node. # class TreeNode: @@ -319,16 +321,25 @@ Python： # self.val = val # self.left = left # self.right = right -# 递归法*前序遍历 class Solution: def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode: - if not root1: return root2 // 如果t1为空，合并之后就应该是t2 - if not root2: return root1 // 如果t2为空，合并之后就应该是t1 - root1.val = root1.val + root2.val //中 - root1.left = self.mergeTrees(root1.left , root2.left) //左 - root1.right = self.mergeTrees(root1.right , root2.right) //右 - return root1 //root1修改了结构和数值 + # 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. + root1.val += root2.val # 中 + root1.left = self.mergeTrees(root1.left, root2.left) #左 + root1.right = self.mergeTrees(root1.right, root2.right) # 右 + + 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. +``` + +**迭代法** +```python # 迭代法-覆盖原来的树 class Solution: def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode: From 84a12f32de997868efcfff08d25c6a161fcc5cff Mon Sep 17 00:00:00 2001 From: Kelvin Date: Sat, 31 Jul 2021 23:10:30 -0400 Subject: [PATCH 27/53] =?UTF-8?q?Update=200617.=E5=90=88=E5=B9=B6=E4=BA=8C?= =?UTF-8?q?=E5=8F=89=E6=A0=91.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 迭代法: 逻辑没有改, 只是改变顺序使之与c++代码逻辑一致. --- problems/0617.合并二叉树.md | 51 ++++++++++++++++++-------------- 1 file changed, 29 insertions(+), 22 deletions(-) diff --git a/problems/0617.合并二叉树.md b/problems/0617.合并二叉树.md index c5eb7594..09d844f3 100644 --- a/problems/0617.合并二叉树.md +++ b/problems/0617.合并二叉树.md @@ -340,31 +340,38 @@ class Solution: **迭代法** ```python -# 迭代法-覆盖原来的树 class Solution: def mergeTrees(self, root1: TreeNode, root2: TreeNode) -> TreeNode: - if not root1: return root2 - if not root2: return root1 - # 迭代,将树2覆盖到树1 - queue1 = [root1] - queue2 = [root2] - root = root1 - while queue1 and queue2: - root1 = queue1.pop(0) - root2 = queue2.pop(0) - root1.val += root2.val - if not root1.left: # 如果树1左儿子不存在，则覆盖后树1的左儿子为树2的左儿子 - root1.left = root2.left - elif root1.left and root2.left: - queue1.append(root1.left) - queue2.append(root2.left) + if not root1: + return root2 + if not root2: + return root1 - if not root1.right: # 同理，处理右儿子 - root1.right = root2.right - elif root1.right and root2.right: - queue1.append(root1.right) - queue2.append(root2.right) - return root + queue = deque() + queue.append(root1) + queue.append(root2) + + while queue: + node1 = queue.popleft() + node2 = queue.popleft() + # 更新queue + # 只有两个节点都有左节点时, 再往queue里面放. + if node1.left and node2.left: + queue.append(node1.left) + queue.append(node2.left) + # 只有两个节点都有右节点时, 再往queue里面放. + if node1.right and node2.right: + queue.append(node1.right) + queue.append(node2.right) + + # 更新当前节点. 同时改变当前节点的左右孩子. + node1.val += node2.val + if not node1.left and node2.left: + node1.left = node2.left + if not node1.right and node2.right: + node1.right = node2.right + + return root1 ``` Go： From 5d1c0883079d6e791804e086e928882a7c091831 Mon Sep 17 00:00:00 2001 From: posper Date: Sun, 1 Aug 2021 11:16:22 +0800 Subject: [PATCH 28/53] =?UTF-8?q?232.=E7=94=A8=E6=A0=88=E5=AE=9E=E7=8E=B0?= =?UTF-8?q?=E9=98=9F=E5=88=97=EF=BC=8C=E4=B9=8B=E5=89=8D=E7=9A=84js?= =?UTF-8?q?=E4=BB=A3=E7=A0=81=E6=94=BE=E5=88=B0=E4=BA=86go=E7=9A=84?= =?UTF-8?q?=E4=BB=A3=E7=A0=81=E5=9D=97=E4=B8=AD?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0232.用栈实现队列.md | 4 +--- 1 file changed, 1 insertion(+), 3 deletions(-) diff --git a/problems/0232.用栈实现队列.md b/problems/0232.用栈实现队列.md index 6890fc2b..32476d6f 100644 --- a/problems/0232.用栈实现队列.md +++ b/problems/0232.用栈实现队列.md @@ -384,7 +384,7 @@ func (this *MyQueue) Peek() int { func (this *MyQueue) Empty() bool { return len(this.stack) == 0 && len(this.back) == 0 } - +``` javaScript: @@ -442,8 +442,6 @@ MyQueue.prototype.empty = function() { ``` - - ----------------------- * 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 7757a60d0ebbe729443be57adccfe3716837dfc9 Mon Sep 17 00:00:00 2001 From: posper Date: Sun, 1 Aug 2021 11:20:13 +0800 Subject: [PATCH 29/53] =?UTF-8?q?34.=E5=9C=A8=E6=8E=92=E5=BA=8F=E6=95=B0?= =?UTF-8?q?=E7=BB=84=E4=B8=AD=E6=9F=A5=E6=89=BE=E5=85=83=E7=B4=A0=E7=9A=84?= =?UTF-8?q?=E7=AC=AC=E4=B8=80=E4=B8=AA=E5=92=8C=E6=9C=80=E5=90=8E=E4=B8=80?= =?UTF-8?q?=E4=B8=AA=E4=BD=8D=E7=BD=AE=EF=BC=8CJava=E7=89=88=E6=9C=AC?= =?UTF-8?q?=E8=A7=A3=E6=B3=952=E6=B3=A8=E9=87=8A=E4=BF=AE=E6=94=B9?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...序数组中查找元素的第一个和最后一个位置.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md index 97b9a9b6..f6b01dae 100644 --- a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md +++ b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md @@ -223,7 +223,7 @@ class Solution { // 解法2 // 1、首先，在 nums 数组中二分查找 target； // 2、如果二分查找失败，则 binarySearch 返回 -1，表明 nums 中没有 target。此时，searchRange 直接返回 {-1, -1}； -// 3、如果二分查找失败，则 binarySearch 返回 nums 中 为 target 的一个下标。然后，通过左右滑动指针，来找到符合题意的区间 +// 3、如果二分查找成功，则 binarySearch 返回 nums 中值为 target 的一个下标。然后，通过左右滑动指针，来找到符合题意的区间 class Solution { public int[] searchRange(int[] nums, int target) { From 173618289d531b74e1a7e04972d88dcb790a9ac0 Mon Sep 17 00:00:00 2001 From: YellowPeaches <326826436@qq.com> Date: Sun, 1 Aug 2021 13:44:57 +0800 Subject: [PATCH 30/53] =?UTF-8?q?=E6=9B=B4=E6=96=B0143.=E9=87=8D=E6=8E=92?= =?UTF-8?q?=E9=93=BE=E8=A1=A8=20=E5=A2=9E=E5=8A=A0Java=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0143.重排链表.md | 49 ++++++++++++++++++++++++++++++++++- 1 file changed, 48 insertions(+), 1 deletion(-) diff --git a/problems/0143.重排链表.md b/problems/0143.重排链表.md index c4e8d8f7..dc0095d3 100644 --- a/problems/0143.重排链表.md +++ b/problems/0143.重排链表.md @@ -1,4 +1,3 @@ -

@@ -63,6 +62,7 @@ public: ## 方法二 把链表放进双向队列，然后通过双向队列一前一后弹出数据，来构造新的链表。这种方法比操作数组容易一些，不用双指针模拟一前一后了 + ```C++ class Solution { public: @@ -176,6 +176,51 @@ public: Java： +```java +public class ReorderList { + public void reorderList(ListNode head) { + ListNode fast = head, slow = head; + //求出中点 + while (fast.next != null && fast.next.next != null) { + slow = slow.next; + fast = fast.next.next; + } + //right就是右半部分 12345 就是45 1234 就是34 + ListNode right = slow.next; + //断开左部分和右部分 + slow.next = null; + //反转右部分 right就是反转后右部分的起点 + right = reverseList(right); + //左部分的起点 + ListNode left = head; + //进行左右部分来回连接 + //这里左部分的节点个数一定大于等于右部分的节点个数 因此只判断right即可 + while (right != null) { + ListNode curLeft = left.next; + left.next = right; + left = curLeft; + + ListNode curRight = right.next; + right.next = left; + right = curRight; + } + } + + public ListNode reverseList(ListNode head) { + ListNode headNode = new ListNode(0); + ListNode cur = head; + ListNode next = null; + while (cur != null) { + next = cur.next; + cur.next = headNode.next; + headNode.next = cur; + cur = next; + } + return headNode.next; + } +} +``` + Python： Go： @@ -183,8 +228,10 @@ Go： JavaScript： ----------------------- + * 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) * 知识星球：[代码随想录](https://mp.weixin.qq.com/s/QVF6upVMSbgvZy8lHZS3CQ) +

From a8109b40141202dd9782f5389f0816c8057999c5 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Sun, 1 Aug 2021 10:15:22 -0400 Subject: [PATCH 31/53] =?UTF-8?q?Update=200700.=E4=BA=8C=E5=8F=89=E6=90=9C?= =?UTF-8?q?=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84=E6=90=9C=E7=B4=A2.md?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 更改代码符合PEP 8 规范. --- problems/0700.二叉搜索树中的搜索.md | 21 +++++++++++--------- 1 file changed, 12 insertions(+), 9 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 d6899ac5..e6eb0358 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 @@ -212,17 +212,20 @@ Python： 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 searchBST(self, root: TreeNode, val: int) -> TreeNode: - if not root or root.val == val: return root //为空或者已经找到都是直接返回root，所以合并了 - if root.val > val: return self.searchBST(root.left,val) //注意一定要加return - else: return self.searchBST(root.right,val) + # 为什么要有返回值: + # 因为搜索到目标节点就要立即return， + # 这样才是找到节点就返回（搜索某一条边），如果不加return，就是遍历整棵树了。 + + if not root or root.val == val: + return root + + if root.val > val: + return self.searchBST(root.left, val) + + if root.val < val: + return self.searchBST(root.right, val) ``` From 2a94d6ca0f413d461537f112cfb8b99855b13cd5 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Sun, 1 Aug 2021 10:17:03 -0400 Subject: [PATCH 32/53] =?UTF-8?q?Update=200700.=E4=BA=8C=E5=8F=89=E6=90=9C?= =?UTF-8?q?=E7=B4=A2=E6=A0=91=E4=B8=AD=E7=9A=84=E6=90=9C=E7=B4=A2.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, 6 insertions(+) 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 e6eb0358..4b02f8af 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 @@ -212,6 +212,12 @@ Python： 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 searchBST(self, root: TreeNode, val: int) -> TreeNode: # 为什么要有返回值: From 3ef151383e3e6aec9b4c6ae14e9dcfa2138f0b8b Mon Sep 17 00:00:00 2001 From: fixme Date: Sun, 1 Aug 2021 23:06:26 +0800 Subject: [PATCH 33/53] =?UTF-8?q?=E3=80=90707=5F=E8=AE=BE=E8=AE=A1?= =?UTF-8?q?=E9=93=BE=E8=A1=A8=E3=80=91=E5=A2=9E=E5=8A=A0C=E5=AE=9E?= =?UTF-8?q?=E7=8E=B0?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0707.设计链表.md | 122 ++++++++++++++++++++++++++++++++++ 1 file changed, 122 insertions(+) diff --git a/problems/0707.设计链表.md b/problems/0707.设计链表.md index 8be28f88..349e2409 100644 --- a/problems/0707.设计链表.md +++ b/problems/0707.设计链表.md @@ -154,7 +154,129 @@ private: ## 其他语言版本 +C: +```C +typedef struct { + int val; + struct MyLinkedList* next; +}MyLinkedList; +/** Initialize your data structure here. */ + +MyLinkedList* myLinkedListCreate() { + //这个题必须用虚拟头指针,参数都是一级指针,头节点确定后没法改指向了!!! + MyLinkedList* head = (MyLinkedList *)malloc(sizeof (MyLinkedList)); + head->next = NULL; + return head; +} + +/** Get the value of the index-th node in the linked list. If the index is invalid, return -1. */ +int myLinkedListGet(MyLinkedList* obj, int index) { + MyLinkedList *cur = obj->next; + for (int i = 0; cur != NULL; i++){ + if (i == index){ + return cur->val; + } + else{ + cur = cur->next; + } + } + return -1; +} + +/** 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. */ +void myLinkedListAddAtHead(MyLinkedList* obj, int val) { + MyLinkedList *nhead = (MyLinkedList *)malloc(sizeof (MyLinkedList)); + nhead->val = val; + nhead->next = obj->next; + obj->next = nhead; + +} + +/** Append a node of value val to the last element of the linked list. */ +void myLinkedListAddAtTail(MyLinkedList* obj, int val) { + MyLinkedList *cur = obj; + while(cur->next != NULL){ + cur = cur->next; + } + MyLinkedList *ntail = (MyLinkedList *)malloc(sizeof (MyLinkedList)); + ntail->val = val; + ntail->next = NULL; + cur->next = ntail; +} + +/** 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. */ +void myLinkedListAddAtIndex(MyLinkedList* obj, int index, int val) { + if (index == 0){ + myLinkedListAddAtHead(obj, val); + return; + } + MyLinkedList *cur = obj->next; + for (int i = 1 ;cur != NULL; i++){ + if (i == index){ + MyLinkedList* newnode = (MyLinkedList *)malloc(sizeof (MyLinkedList)); + newnode->val = val; + newnode->next = cur->next; + cur->next = newnode; + return; + } + else{ + cur = cur->next; + } + } +} + +/** Delete the index-th node in the linked list, if the index is valid. */ +void myLinkedListDeleteAtIndex(MyLinkedList* obj, int index) { + if (index == 0){ + MyLinkedList *tmp = obj->next; + if (tmp != NULL){ + obj->next = tmp->next; + free(tmp) + } + return; + } + MyLinkedList *cur = obj->next; + for (int i = 1 ;cur != NULL && cur->next != NULL; i++){ + if (i == index){ + MyLinkedList *tmp = cur->next; + if (tmp != NULL) { + cur->next = tmp->next; + free(tmp); + } + return; + } + else{ + cur = cur->next; + } + } + +} + +void myLinkedListFree(MyLinkedList* obj) { + while(obj != NULL){ + MyLinkedList *tmp = obj; + obj = obj->next; + free(tmp); + } +} + +/** + * Your MyLinkedList struct will be instantiated and called as such: + * MyLinkedList* obj = myLinkedListCreate(); + * int param_1 = myLinkedListGet(obj, index); + + * myLinkedListAddAtHead(obj, val); + + * myLinkedListAddAtTail(obj, val); + + * myLinkedListAddAtIndex(obj, index, val); + + * myLinkedListDeleteAtIndex(obj, index); + + * myLinkedListFree(obj); +*/ +``` Java： ```Java From b87d619cdd1ab886055d94cb4b878f550fe51ea7 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Sun, 1 Aug 2021 13:54:51 -0400 Subject: [PATCH 34/53] =?UTF-8?q?Update=200098.=E9=AA=8C=E8=AF=81=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 更改递归部分代码 1. 更新上一版代码格式 使其更符合PEP8. 2. 把代码细节都暴露出来, 逻辑更加清晰. --- problems/0098.验证二叉搜索树.md | 34 ++++++++++++++++++-------- 1 file changed, 24 insertions(+), 10 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 248d10f1..a271b977 100644 --- a/problems/0098.验证二叉搜索树.md +++ b/problems/0098.验证二叉搜索树.md @@ -337,6 +337,8 @@ class Solution { ``` Python： + +**递归** - 利用BST中序遍历特性 ```python # Definition for a binary tree node. # class TreeNode: @@ -344,18 +346,30 @@ Python： # self.val = val # self.left = left # self.right = right -# 递归法 class Solution: def isValidBST(self, root: TreeNode) -> bool: - res = [] //把二叉搜索树按中序遍历写成list - def buildalist(root): - if not root: return - buildalist(root.left) //左 - res.append(root.val) //中 - buildalist(root.right) //右 - return res - buildalist(root) - return res == sorted(res) and len(set(res)) == len(res) //检查list里的数有没有重复元素，以及是否按从小到大排列 + # 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]: + return False + return True + + __traverse(root) + res = __is_sorted(candidate_list) + + return res # 简单递归法 class Solution: From 6178dff832cebd336c596ae40ae941ab28cd42c2 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Sun, 1 Aug 2021 13:57:12 -0400 Subject: [PATCH 35/53] =?UTF-8?q?Update=200098.=E9=AA=8C=E8=AF=81=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 增加comment --- problems/0098.验证二叉搜索树.md | 9 ++++++--- 1 file changed, 6 insertions(+), 3 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 a271b977..cf8651f3 100644 --- a/problems/0098.验证二叉搜索树.md +++ b/problems/0098.验证二叉搜索树.md @@ -338,7 +338,7 @@ class Solution { Python： -**递归** - 利用BST中序遍历特性 +**递归** - 利用BST中序遍历特性,把树"压缩"成数组 ```python # Definition for a binary tree node. # class TreeNode: @@ -362,7 +362,7 @@ class Solution: def __is_sorted(nums: list) -> bool: for i in range(1, len(nums)): - if nums[i] <= nums[i - 1]: + 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 @@ -370,8 +370,11 @@ class Solution: res = __is_sorted(candidate_list) return res +``` -# 简单递归法 +**递归** - + +```python class Solution: def isValidBST(self, root: TreeNode) -> bool: def isBST(root, min_val, max_val): From 59fdafc7a116ca694b19ba78721bdc416cda76e8 Mon Sep 17 00:00:00 2001 From: Jack <965555169@qq.com> Date: Mon, 2 Aug 2021 08:58:19 +0800 Subject: [PATCH 36/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A053.=E6=9C=80=E5=A4=A7?= =?UTF-8?q?=E5=AD=90=E5=BA=8F=E5=92=8C(=E5=8A=A8=E6=80=81=E8=A7=84?= =?UTF-8?q?=E5=88=92)-JavaScript=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../0053.最大子序和（动态规划）.md | 17 +++++++++++++++++ 1 file changed, 17 insertions(+) diff --git a/problems/0053.最大子序和（动态规划）.md b/problems/0053.最大子序和（动态规划）.md index ada378b2..20a8c66c 100644 --- a/problems/0053.最大子序和（动态规划）.md +++ b/problems/0053.最大子序和（动态规划）.md @@ -168,6 +168,23 @@ func max(a,b int) int{ } ``` +JavaScript： + +```javascript +const maxSubArray = nums => { + // 数组长度，dp初始化 + const [len, dp] = [nums.length, [nums[0]]]; + // 最大值初始化为dp[0] + let max = dp[0]; + for (let i = 1; i < len; i++) { + dp[i] = Math.max(dp[i - 1] + nums[i], nums[i]); + // 更新最大值 + max = Math.max(max, dp[i]); + } + return max; +}; +``` + ----------------------- From 9a87b13fc3f87c1b46310d1c8da92b7b8ca8d078 Mon Sep 17 00:00:00 2001 From: Jack <965555169@qq.com> Date: Mon, 2 Aug 2021 09:16:10 +0800 Subject: [PATCH 37/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A0392.=E5=88=A4=E6=96=AD?= =?UTF-8?q?=E5=AD=90=E5=BA=8F=E5=88=97(=E5=8A=A8=E6=80=81=E8=A7=84?= =?UTF-8?q?=E5=88=92)-JavaScript=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0392.判断子序列.md | 25 ++++++++++++++++++++++++- 1 file changed, 24 insertions(+), 1 deletion(-) diff --git a/problems/0392.判断子序列.md b/problems/0392.判断子序列.md index d97d2684..09ee2c4d 100644 --- a/problems/0392.判断子序列.md +++ b/problems/0392.判断子序列.md @@ -180,7 +180,30 @@ class Solution: return False ``` -Go： +JavaScript： + +```javascript +const isSubsequence = (s, t) => { + // s、t的长度 + const [m, n] = [s.length, t.length]; + // dp全初始化为0 + const dp = new Array(m + 1).fill(0).map(x => new Array(n + 1).fill(0)); + for (let i = 1; i <= m; i++) { + for (let j = 1; j <= n; j++) { + // 更新dp[i][j]，两种情况 + if (s[i - 1] === t[j - 1]) { + dp[i][j] = dp[i - 1][j - 1] + 1; + } else { + dp[i][j] = dp[i][j - 1]; + } + } + } + // 遍历结束，判断dp右下角的数是否等于s的长度 + return dp[m][n] === m ? true : false; +}; +``` + + From 0fdede8c944b5052160370b3fe1241810021cfa3 Mon Sep 17 00:00:00 2001 From: Kelvin Date: Sun, 1 Aug 2021 22:53:23 -0400 Subject: [PATCH 38/53] =?UTF-8?q?Update=200098.=E9=AA=8C=E8=AF=81=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 将上一版本的代码改成跟题解思路相同. --- problems/0098.验证二叉搜索树.md | 26 +++++++++++++++++++------- 1 file changed, 19 insertions(+), 7 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 cf8651f3..a7634849 100644 --- a/problems/0098.验证二叉搜索树.md +++ b/problems/0098.验证二叉搜索树.md @@ -372,18 +372,30 @@ class Solution: return res ``` -**递归** - +**递归** - 标准做法 ```python class Solution: def isValidBST(self, root: TreeNode) -> bool: - def isBST(root, min_val, max_val): - if not root: return True - if root.val >= max_val or root.val <= min_val: + # 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: return False - return isBST(root.left, min_val, root.val) and isBST(root.right, root.val, max_val) - return isBST(root, float("-inf"), float("inf")) - + is_right_valid = __isValidBST(root.right) + + return is_left_valid and is_right_valid + return __isValidBST(root) +``` +``` # 迭代-中序遍历 class Solution: def isValidBST(self, root: TreeNode) -> bool: From 152f7cc61a8e5891374b14c0ac998995fe3da767 Mon Sep 17 00:00:00 2001 From: Jack <965555169@qq.com> Date: Tue, 3 Aug 2021 08:45:05 +0800 Subject: [PATCH 39/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A01035.=E4=B8=8D=E7=9B=B8?= =?UTF-8?q?=E4=BA=A4=E7=9A=84=E7=BA=BF-JavaScript=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1035.不相交的线.md | 23 ++++++++++++++++++++++- 1 file changed, 22 insertions(+), 1 deletion(-) diff --git a/problems/1035.不相交的线.md b/problems/1035.不相交的线.md index cb74ef75..3a0e6a68 100644 --- a/problems/1035.不相交的线.md +++ b/problems/1035.不相交的线.md @@ -109,7 +109,28 @@ class Solution: return dp[-1][-1] ``` -Go： +JavaScript： + +```javascript +const maxUncrossedLines = (nums1, nums2) => { + // 两个数组长度 + const [m, n] = [nums1.length, nums2.length]; + // 创建dp数组并都初始化为0 + const dp = new Array(m + 1).fill(0).map(x => new Array(n + 1).fill(0)); + for (let i = 1; i <= m; i++) { + for (let j = 1; j <= n; j++) { + // 根据两种情况更新dp[i][j] + if (nums1[i - 1] === nums2[j - 1]) { + dp[i][j] = dp[i - 1][j - 1] + 1; + } else { + dp[i][j] = Math.max(dp[i - 1][j], dp[i][j - 1]); + } + } + } + // 返回dp数组中右下角的元素 + return dp[m][n]; +}; +``` From 7cc60a02a7c62df1784ad9deca39cc7822f0e6e7 Mon Sep 17 00:00:00 2001 From: ltinyho Date: Tue, 3 Aug 2021 10:34:24 +0800 Subject: [PATCH 40/53] =?UTF-8?q?1002.=E6=9F=A5=E6=89=BE=E5=B8=B8=E7=94=A8?= =?UTF-8?q?=E5=AD=97=E7=AC=A6-JavaScript=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/1002.查找常用字符.md | 40 +++++++++++++++++++++++++++++ 1 file changed, 40 insertions(+) diff --git a/problems/1002.查找常用字符.md b/problems/1002.查找常用字符.md index 875340c4..e3014d00 100644 --- a/problems/1002.查找常用字符.md +++ b/problems/1002.查找常用字符.md @@ -169,6 +169,46 @@ class Solution { } } ``` +javaScript +```js +var commonChars = function (words) { + let res = [] + let size = 26 + let firstHash = new Array(size) + for (let i = 0; i < size; i++) { // 初始化 hash 数组 + firstHash[i] = 0 + } + + let a = "a".charCodeAt() + let firstWord = words[0] + for (let i = 0; i < firstWord.length; i++) { // 第 0 个单词的统计 + let idx = firstWord[i].charCodeAt() + firstHash[idx - a] += 1 + } + + for (let i = 1; i < words.length; i++) { // 1-n 个单词统计 + let otherHash = new Array(size) + for (let i = 0; i < size; i++) { // 初始化 hash 数组 + otherHash[i] = 0 + } + + for (let j = 0; j < words[i].length; j++) { + let idx = words[i][j].charCodeAt() + otherHash[idx - a] += 1 + } + for (let i = 0; i < size; i++) { + firstHash[i] = Math.min(firstHash[i], otherHash[i]) + } + } + for (let i = 0; i < size; i++) { + while (firstHash[i] > 0) { + res.push(String.fromCharCode(i + a)) + firstHash[i]-- + } + } + return res +}; +``` ----------------------- * 作者微信：[程序员Carl](https://mp.weixin.qq.com/s/b66DFkOp8OOxdZC_xLZxfw) From 4068040965153cbcbdd970be6abf5fa31be517f0 Mon Sep 17 00:00:00 2001 From: ltinyho Date: Tue, 3 Aug 2021 14:43:44 +0800 Subject: [PATCH 41/53] =?UTF-8?q?0018.=E5=9B=9B=E6=95=B0=E4=B9=8B=E5=92=8C?= =?UTF-8?q?-golang?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0018.四数之和.md | 48 +++++++++++++++++++++++++++++++++++ 1 file changed, 48 insertions(+) diff --git a/problems/0018.四数之和.md b/problems/0018.四数之和.md index 0caf12be..4a8fa947 100644 --- a/problems/0018.四数之和.md +++ b/problems/0018.四数之和.md @@ -201,6 +201,54 @@ class Solution(object): ``` Go： +```go +func fourSum(nums []int, target int) [][]int { + if len(nums) < 4 { + return nil + } + sort.Ints(nums) + var res [][]int + for i := 0; i < len(nums)-3; i++ { + n1 := nums[i] + // if n1 > target { // 不能这样写,因为可能是负数 + // break + // } + if i > 0 && n1 == nums[i-1] { + continue + } + for j := i + 1; j < len(nums)-2; j++ { + n2 := nums[j] + if j > i+1 && n2 == nums[j-1] { + continue + } + l := j + 1 + r := len(nums) - 1 + for l < r { + n3 := nums[l] + n4 := nums[r] + sum := n1 + n2 + n3 + n4 + if sum < target { + l++ + } else if sum > target { + r-- + } else { + res = append(res, []int{n1, n2, n3, n4}) + for l < r && n3 == nums[l+1] { // 去重 + l++ + } + for l < r && n4 == nums[r-1] { // 去重 + r-- + } + // 找到答案时,双指针同时靠近 + r-- + l++ + } + } + } + } + return res +} +``` javaScript: From 7ff7afcfaac673ba846cc5908aa68ee13d811fe0 Mon Sep 17 00:00:00 2001 From: YellowPeaches <326826436@qq.com> Date: Tue, 3 Aug 2021 16:25:05 +0800 Subject: [PATCH 42/53] =?UTF-8?q?update=200100=E7=9B=B8=E5=90=8C=E7=9A=84?= =?UTF-8?q?=E6=A0=91java=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0100.相同的树.md | 16 ++++++++++++++++ 1 file changed, 16 insertions(+) diff --git a/problems/0100.相同的树.md b/problems/0100.相同的树.md index b536ffe3..c40f79fd 100644 --- a/problems/0100.相同的树.md +++ b/problems/0100.相同的树.md @@ -172,6 +172,22 @@ public: Java： +```java +class Solution { + public boolean isSameTree(TreeNode p, TreeNode q) { + if (p == null && q == null) { + return true; + } else if (q == null || p == null) { + return false; + } else if (q.val != p.val) { + return false; + } else { + return isSameTree(q.left, p.left) && isSameTree(q.right, p.right); + } + } +} + +``` Python： Go： From 60799a2f446b0b54bb6dcf29ef302ecf9dfeb3e6 Mon Sep 17 00:00:00 2001 From: posper Date: Tue, 3 Aug 2021 20:53:53 +0800 Subject: [PATCH 43/53] =?UTF-8?q?=E4=BA=8C=E5=8F=89=E6=A0=91=E5=B1=82?= =?UTF-8?q?=E5=BA=8F=E9=81=8D=E5=8E=86=20116.=E5=A1=AB=E5=85=85=E6=AF=8F?= =?UTF-8?q?=E4=B8=AA=E8=8A=82=E7=82=B9=E7=9A=84=E4=B8=8B=E4=B8=80=E4=B8=AA?= =?UTF-8?q?=E5=8F=B3=E4=BE=A7=E8=8A=82=E7=82=B9=E6=8C=87=E9=92=88=20?= =?UTF-8?q?=E6=B7=BB=E5=8A=A0Java=E7=89=88=E6=9C=AC?= 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 | 38 +++++++++++++++++++++++ 1 file changed, 38 insertions(+) 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 8be3ac47..7d459ad9 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 @@ -1254,6 +1254,44 @@ func connect(root *Node) *Node { } ``` +Java 代码： + +```java +// 二叉树之层次遍历 +class Solution { + public Node connect(Node root) { + Queue queue = new LinkedList<>(); + if (root != null) { + queue.add(root); + } + while (!queue.isEmpty()) { + int size = queue.size(); + Node node = null; + Node nodePre = null; + + for (int i = 0; i < size; i++) { + if (i == 0) { + nodePre = queue.poll(); // 取出本层头一个节点 + node = nodePre; + } else { + node = queue.poll(); + nodePre.next = node; // 本层前一个节点 next 指向当前节点 + nodePre = nodePre.next; + } + if (node.left != null) { + queue.add(node.left); + } + if (node.right != null) { + queue.add(node.right); + } + } + nodePre.next = null; // 本层最后一个节点 next 指向 null + } + return root; + } +} +``` + ## 117.填充每个节点的下一个右侧节点指针II 题目地址：https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node-ii/ From c62d6255013161bfba38ebf70e38cca410dba15c Mon Sep 17 00:00:00 2001 From: ironartisan Date: Tue, 3 Aug 2021 21:58:23 +0800 Subject: [PATCH 44/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200034.=E5=9C=A8?= =?UTF-8?q?=E6=8E=92=E5=BA=8F=E6=95=B0=E7=BB=84=E4=B8=AD=E6=9F=A5=E6=89=BE?= =?UTF-8?q?=E5=85=83=E7=B4=A0=E7=9A=84=E7=AC=AC=E4=B8=80=E4=B8=AA=E5=92=8C?= =?UTF-8?q?=E6=9C=80=E5=90=8E=E4=B8=80=E4=B8=AA=E4=BD=8D=E7=BD=AEpython3?= =?UTF-8?q?=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- ...元素的第一个和最后一个位置.md | 113 +++++++++++++++++- 1 file changed, 112 insertions(+), 1 deletion(-) diff --git a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md index f6b01dae..ced4d4ac 100644 --- a/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md +++ b/problems/0034.在排序数组中查找元素的第一个和最后一个位置.md @@ -50,7 +50,7 @@ 接下来，在去寻找左边界，和右边界了。 -采用二分法来取寻找左右边界，为了让代码清晰，我分别写两个二分来寻找左边界和右边界。 +采用二分法来去寻找左右边界，为了让代码清晰，我分别写两个二分来寻找左边界和右边界。 **刚刚接触二分搜索的同学不建议上来就像如果用一个二分来查找左右边界，很容易把自己绕进去，建议扎扎实实的写两个二分分别找左边界和右边界** @@ -275,6 +275,117 @@ class Solution { ## Python ```python +class Solution: + def searchRange(self, nums: List[int], target: int) -> List[int]: + def getRightBorder(nums:List[int], target:int) -> int: + left, right = 0, len(nums)-1 + rightBoder = -2 # 记录一下rightBorder没有被赋值的情况 + while left <= right: + middle = left + (right-left) // 2 + if nums[middle] > target: + right = middle - 1 + else: # 寻找右边界，nums[middle] == target的时候更新left + left = middle + 1 + rightBoder = left + + return rightBoder + + def getLeftBorder(nums:List[int], target:int) -> int: + left, right = 0, len(nums)-1 + leftBoder = -2 # 记录一下leftBorder没有被赋值的情况 + while left <= right: + middle = left + (right-left) // 2 + if nums[middle] >= target: # 寻找左边界，nums[middle] == target的时候更新right + right = middle - 1; + leftBoder = right; + else: + left = middle + 1 + return leftBoder + leftBoder = getLeftBorder(nums, target) + rightBoder = getRightBorder(nums, target) + # 情况一 + if leftBoder == -2 or rightBoder == -2: return [-1, -1] + # 情况三 + if rightBoder -leftBoder >1: return [leftBoder + 1, rightBoder - 1] + # 情况二 + return [-1, -1] +``` +```python +# 解法2 +# 1、首先，在 nums 数组中二分查找 target； +# 2、如果二分查找失败，则 binarySearch 返回 -1，表明 nums 中没有 target。此时，searchRange 直接返回 {-1, -1}； +# 3、如果二分查找成功，则 binarySearch 返回 nums 中值为 target 的一个下标。然后，通过左右滑动指针，来找到符合题意的区间 +class Solution: + def searchRange(self, nums: List[int], target: int) -> List[int]: + def binarySearch(nums:List[int], target:int) -> int: + left, right = 0, len(nums)-1 + while left<=right: # 不变量：左闭右闭区间 + middle = left + (right-left) // 2 + if nums[middle] > target: + right = middle - 1 + elif nums[middle] < target: + left = middle + 1 + else: + return middle + return -1 + index = binarySearch(nums, target) + if index == -1:return [-1, -1] # nums 中不存在 target，直接返回 {-1, -1} + # nums 中存在 targe，则左右滑动指针，来找到符合题意的区间 + left, right = index, index + # 向左滑动，找左边界 + while left -1 >=0 and nums[left - 1] == target: left -=1 + # 向右滑动，找右边界 + while right+1 < len(nums) and nums[right + 1] == target: right +=1 + return [left, right] +``` +```python +# 解法3 +# 1、首先，在 nums 数组中二分查找得到第一个大于等于 target的下标（左边界）与第一个大于target的下标（右边界）； +# 2、如果左边界<= 右边界，则返回 [左边界, 右边界]。否则返回[-1, -1] +class Solution: + def searchRange(self, nums: List[int], target: int) -> List[int]: + def binarySearch(nums:List[int], target:int, lower:bool) -> int: + left, right = 0, len(nums)-1 + ans = len(nums) + while left<=right: # 不变量：左闭右闭区间 + middle = left + (right-left) //2 + # lower为True，执行前半部分，找到第一个大于等于 target的下标 ，否则找到第一个大于target的下标 + if nums[middle] > target or (lower and nums[middle] >= target): + right = middle - 1 + ans = middle + else: + left = middle + 1 + return ans + + leftBorder = binarySearch(nums, target, True) # 搜索左边界 + rightBorder = binarySearch(nums, target, False) -1 # 搜索右边界 + if leftBorder<= rightBorder and rightBorder< len(nums) and nums[leftBorder] == target and nums[rightBorder] == target: + return [leftBorder, rightBorder] + return [-1, -1] +``` + +```python +# 解法4 +# 1、首先，在 nums 数组中二分查找得到第一个大于等于 target的下标leftBorder； +# 2、在 nums 数组中二分查找得到第一个大于等于 target+1的下标， 减1则得到rightBorder； +# 3、如果开始位置在数组的右边或者不存在target，则返回[-1, -1] 。否则返回[leftBorder, rightBorder] +class Solution: + def searchRange(self, nums: List[int], target: int) -> List[int]: + def binarySearch(nums:List[int], target:int) -> int: + left, right = 0, len(nums)-1 + while left<=right: # 不变量：左闭右闭区间 + middle = left + (right-left) //2 + if nums[middle] >= target: + right = middle - 1 + else: + left = middle + 1 + return left # 若存在target，则返回第一个等于target的值 + + leftBorder = binarySearch(nums, target) # 搜索左边界 + rightBorder = binarySearch(nums, target+1) -1 # 搜索右边界 + if leftBorder == len(nums) or nums[leftBorder]!= target: # 情况一和情况二 + return [-1, -1] + return [leftBorder, rightBorder] ``` ## Go From 26a07b9bf1597ceda73c4a616aea29395f2c0341 Mon Sep 17 00:00:00 2001 From: fixme Date: Tue, 3 Aug 2021 22:10:15 +0800 Subject: [PATCH 45/53] =?UTF-8?q?=E3=80=9024.=20=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=E3=80=91C=E5=AE=9E=E7=8E=B0(=E5=8F=8C=E6=8C=87?= =?UTF-8?q?=E9=92=88)?= 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 | 28 +++++++++++++++++++ 1 file changed, 28 insertions(+) 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 66e149e6..b285b2d4 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 @@ -86,6 +86,34 @@ public: ## 其他语言版本 +C: +``` +/** + * Definition for singly-linked list. + * struct ListNode { + * int val; + * struct ListNode *next; + * }; + */ + + +struct ListNode* swapPairs(struct ListNode* head){ + //使用双指针避免使用中间变量 + typedef struct ListNode ListNode; + ListNode *fakehead = (ListNode *)malloc(sizeof(ListNode)); + fakehead->next = head; + ListNode* right = fakehead->next; + ListNode* left = fakehead; + while(left && right && right->next ){ + left->next = right->next; + right->next = left->next->next; + left->next->next = right; + left = right; + right = left->next; + } + return fakehead->next; +} +``` Java： From 098bb7b262e57d7c0ae9a51ab484665577835b4f Mon Sep 17 00:00:00 2001 From: Jack <965555169@qq.com> Date: Wed, 4 Aug 2021 12:08:00 +0800 Subject: [PATCH 46/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A0198.=E6=89=93=E5=AE=B6?= =?UTF-8?q?=E5=8A=AB=E8=88=8D-JavaScript=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0198.打家劫舍.md | 18 +++++++++++++++++- 1 file changed, 17 insertions(+), 1 deletion(-) diff --git a/problems/0198.打家劫舍.md b/problems/0198.打家劫舍.md index 63a68c36..5b5863a4 100644 --- a/problems/0198.打家劫舍.md +++ b/problems/0198.打家劫舍.md @@ -25,7 +25,7 @@ 输出：12 解释：偷窃 1 号房屋 (金额 = 2), 偷窃 3 号房屋 (金额 = 9)，接着偷窃 5 号房屋 (金额 = 1)。   偷窃到的最高金额 = 2 + 9 + 1 = 12 。 -  + 提示： @@ -175,6 +175,22 @@ func max(a, b int) int { } ``` +JavaScript： + +```javascript +const rob = nums => { + // 数组长度 + const len = nums.length; + // dp数组初始化 + const dp = [nums[0], Math.max(nums[0], nums[1])]; + // 从下标2开始遍历 + for (let i = 2; i < len; i++) { + dp[i] = Math.max(dp[i - 2] + nums[i], dp[i - 1]); + } + return dp[len - 1]; +}; +``` + From 3d880e9ca67afe1b1a2cb19b8d8cba27c1544875 Mon Sep 17 00:00:00 2001 From: SwordsmanYao Date: Wed, 4 Aug 2021 16:26:02 +0800 Subject: [PATCH 47/53] =?UTF-8?q?[0347.=E5=89=8DK=E4=B8=AA=E9=AB=98?= =?UTF-8?q?=E9=A2=91=E5=85=83=E7=B4=A0]=20=E6=B7=BB=E5=8A=A0js=E7=89=88?= =?UTF-8?q?=E6=9C=AC=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0347.前K个高频元素.md | 95 ++++++++++++++++++++++++++++ 1 file changed, 95 insertions(+) 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 6b07c258..161df1ad 100644 --- a/problems/0347.前K个高频元素.md +++ b/problems/0347.前K个高频元素.md @@ -190,6 +190,101 @@ class Solution: Go： +javaScript: +```js +/** + * @param {number[]} nums + * @param {number} k + * @return {number[]} + */ +var topKFrequent = function(nums, k) { + const map = new Map(); + + for(const num of nums) { + map.set(num, (map.get(num) || 0) + 1); + } + + // 创建小顶堆 + const priorityQueue = new PriorityQueue((a, b) => a[1] - b[1]); + + // entry 是一个长度为2的数组，0位置存储key，1位置存储value + for (const entry of map.entries()) { + priorityQueue.push(entry); + if (priorityQueue.size() > k) { + priorityQueue.pop(); + } + } + + const ret = []; + + for(let i = priorityQueue.size() - 1; i >= 0; i--) { + ret[i] = priorityQueue.pop()[0]; + } + + return ret; +}; + + +function PriorityQueue(compareFn) { + this.compareFn = compareFn; + this.queue = []; +} + +// 添加 +PriorityQueue.prototype.push = function(item) { + this.queue.push(item); + let index = this.queue.length - 1; + let parent = Math.floor((index - 1) / 2); + // 上浮 + while(parent >= 0 && this.compare(parent, index) > 0) { + // 交换 + [this.queue[index], this.queue[parent]] = [this.queue[parent], this.queue[index]]; + index = parent; + parent = Math.floor((index - 1) / 2); + } +} + +// 获取堆顶元素并移除 +PriorityQueue.prototype.pop = function() { + const ret = this.queue[0]; + + // 把最后一个节点移到堆顶 + this.queue[0] = this.queue.pop(); + + let index = 0; + // 左子节点下标，left + 1 就是右子节点下标 + let left = 1; + let selectedChild = this.compare(left, left + 1) > 0 ? left + 1 : left; + + // 下沉 + while(selectedChild !== undefined && this.compare(index, selectedChild) > 0) { + // 交换 + [this.queue[index], this.queue[selectedChild]] = [this.queue[selectedChild], this.queue[index]]; + index = selectedChild; + left = 2 * index + 1; + selectedChild = this.compare(left, left + 1) > 0 ? left + 1 : left; + } + + return ret; +} + +PriorityQueue.prototype.size = function() { + return this.queue.length; +} + +// 使用传入的 compareFn 比较两个位置的元素 +PriorityQueue.prototype.compare = function(index1, index2) { + if (this.queue[index1] === undefined) { + return 1; + } + if (this.queue[index2] === undefined) { + return -1; + } + + return this.compareFn(this.queue[index1], this.queue[index2]); +} +``` + ----------------------- From e222588496db4a8f3a5a6a6fa9550dc96c9b335f Mon Sep 17 00:00:00 2001 From: Kelvin Date: Wed, 4 Aug 2021 10:01:52 -0400 Subject: [PATCH 48/53] =?UTF-8?q?=E6=9B=B4=E6=96=B0701.=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=20Python3=20=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 1. 增加题解中提及到的递归方法 - 无返回值 2. 增加题解中提及到的迭代方法 --- .../0701.二叉搜索树中的插入操作.md | 58 ++++++++++++++++++- 1 file changed, 57 insertions(+), 1 deletion(-) 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 61027453..1ec80da1 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 @@ -255,7 +255,7 @@ class Solution { Python： -递归法 +**递归法** - 有返回值 ```python class Solution: @@ -268,7 +268,63 @@ class Solution: root.left = self.insertIntoBST(root.left, val) # 递归创建左子树 return root ``` +**递归法** - 无返回值 +```python +class Solution: + def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode: + if not root: + 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 +``` +**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 + cur = root + + # 用while循环不断地找新节点的parent + while cur: + if cur.val < val: + parent = cur + cur = cur.right + elif cur.val > val: + parent = cur + 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: + parent.left = TreeNode(val) + else: + parent.right = TreeNode(val) + + return root + +``` Go： From a0824a84ec24bec4037087ba7084861885463b8f Mon Sep 17 00:00:00 2001 From: fusunx <1102654482@qq.com> Date: Thu, 5 Aug 2021 08:02:52 +0800 Subject: [PATCH 49/53] =?UTF-8?q?=E8=83=8C=E5=8C=85=E9=97=AE=E9=A2=98?= =?UTF-8?q?=E7=90=86=E8=AE=BA=E5=9F=BA=E7=A1=80=E5=AE=8C=E5=85=A8=E8=83=8C?= =?UTF-8?q?=E5=8C=85=20Javascript?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- .../背包问题理论基础完全背包.md | 31 +++++++++++++++++++ 1 file changed, 31 insertions(+) diff --git a/problems/背包问题理论基础完全背包.md b/problems/背包问题理论基础完全背包.md index a5a708cf..2cbc7bb8 100644 --- a/problems/背包问题理论基础完全背包.md +++ b/problems/背包问题理论基础完全背包.md @@ -272,7 +272,38 @@ func main() { fmt.Println(test_CompletePack2(weight, price, 4)) } ``` +Javascript: +```Javascript +// 先遍历物品，再遍历背包容量 +function test_completePack1() { + let weight = [1, 3, 5] + let value = [15, 20, 30] + let bagWeight = 4 + let dp = new Array(bagWeight + 1).fill(0) + for(let i = 0; i <= weight.length; i++) { + for(let j = weight[i]; j <= bagWeight; j++) { + dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]) + } + } + console.log(dp) +} +// 先遍历背包容量，再遍历物品 +function test_completePack2() { + let weight = [1, 3, 5] + let value = [15, 20, 30] + let bagWeight = 4 + let dp = new Array(bagWeight + 1).fill(0) + for(let j = 0; j <= bagWeight; j++) { + for(let i = 0; i < weight.length; i++) { + if (j >= weight[i]) { + dp[j] = Math.max(dp[j], dp[j - weight[i]] + value[i]) + } + } + } + console.log(2, dp); +} +``` ----------------------- From 993b0795f5af6d9498ee98707eb780ec0c741802 Mon Sep 17 00:00:00 2001 From: ironartisan Date: Thu, 5 Aug 2021 10:11:00 +0800 Subject: [PATCH 50/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200100.=E7=9B=B8?= =?UTF-8?q?=E5=90=8C=E7=9A=84=E6=A0=91=20java=E5=92=8Cpython3=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/0100.相同的树.md | 84 +++++++++++++++++++++++++++-------- 1 file changed, 66 insertions(+), 18 deletions(-) diff --git a/problems/0100.相同的树.md b/problems/0100.相同的树.md index c40f79fd..31a68692 100644 --- a/problems/0100.相同的树.md +++ b/problems/0100.相同的树.md @@ -139,29 +139,29 @@ public: ## 迭代法 ```C++ -lass Solution { +class Solution { public: bool isSameTree(TreeNode* p, TreeNode* q) { if (p == NULL && q == NULL) return true; if (p == NULL || q == NULL) return false; queue que; - que.push(p); // - que.push(q); // + que.push(p); // 添加根节点p + que.push(q); // 添加根节点q while (!que.empty()) { // TreeNode* leftNode = que.front(); que.pop(); TreeNode* rightNode = que.front(); que.pop(); - if (!leftNode && !rightNode) { // + if (!leftNode && !rightNode) { // 若p的节点与q的节点都为空 continue; } - // + // 若p的节点与q的节点有一个为空或p的节点的值与q节点不同 if ((!leftNode || !rightNode || (leftNode->val != rightNode->val))) { return false; } - que.push(leftNode->left); // - que.push(rightNode->left); // - que.push(leftNode->right); // - que.push(rightNode->right); // + que.push(leftNode->left); // 添加p节点的左子树节点 + que.push(rightNode->left); // 添加q节点的左子树节点点 + que.push(leftNode->right); // 添加p节点的右子树节点 + que.push(rightNode->right); // 添加q节点的右子树节点 } return true; } @@ -173,23 +173,71 @@ public: Java： ```java +// 递归法 class Solution { public boolean isSameTree(TreeNode p, TreeNode q) { - if (p == null && q == null) { - return true; - } else if (q == null || p == null) { - return false; - } else if (q.val != p.val) { - return false; - } else { - return isSameTree(q.left, p.left) && isSameTree(q.right, p.right); - } + if (p == null && q == null) return true; + else if (q == null || p == null) return false; + else if (q.val != p.val) return false; + return isSameTree(q.left, p.left) && isSameTree(q.right, p.right); } } +``` +```java +// 迭代法 +class Solution { + public boolean isSameTree(TreeNode p, TreeNode q) { + if(p == null && q == null) return true; + if(p == null || q == null) return false; + Queue que= new LinkedList(); + que.offer(p); + que.offer(q); + while(!que.isEmpty()){ + TreeNode leftNode = que.poll(); + TreeNode rightNode = que.poll(); + if(leftNode == null && rightNode == null) continue; + if(leftNode == null || rightNode== null || leftNode.val != rightNode.val) return false; + que.offer(leftNode.left); + que.offer(rightNode.left); + que.offer(leftNode.right); + que.offer(rightNode.right); + } + return true; + } +} ``` Python： +```python +# 递归法 +class Solution: + def isSameTree(self, p: TreeNode, q: TreeNode) -> bool: + if not p and not q: return True + elif not p or not q: return False + elif p.val != q.val: return False + return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right) +``` +```python +# 迭代法 +class Solution: + def isSameTree(self, p: TreeNode, q: TreeNode) -> bool: + if not p and not q: return True + if not p or not q: return False + que = collections.deque() + que.append(p) + que.append(q) + while que: + leftNode = que.popleft() + rightNode = que.popleft() + if not leftNode and not rightNode: continue + if not leftNode or not rightNode or leftNode.val != rightNode.val: return False + que.append(leftNode.left) + que.append(rightNode.left) + que.append(leftNode.right) + que.append(rightNode.right) + return True +``` Go： JavaScript： From 5865faa73a9a33fab1e199bbce96c29f508a378d Mon Sep 17 00:00:00 2001 From: Jack <965555169@qq.com> Date: Thu, 5 Aug 2021 10:36:48 +0800 Subject: [PATCH 51/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A0337.=E6=89=93=E5=AE=B6?= =?UTF-8?q?=E5=8A=AB=E8=88=8D=20III-=E5=8A=A8=E6=80=81=E8=A7=84=E5=88=92-J?= =?UTF-8?q?avaScript=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0337.打家劫舍III.md | 27 ++++++++++++++++++++++++++- 1 file changed, 26 insertions(+), 1 deletion(-) diff --git a/problems/0337.打家劫舍III.md b/problems/0337.打家劫舍III.md index 197841e9..780e4aef 100644 --- a/problems/0337.打家劫舍III.md +++ b/problems/0337.打家劫舍III.md @@ -368,7 +368,32 @@ class Solution: return (val1, val2) ``` -Go： +JavaScript： + +> 动态规划 + +```javascript +const rob = root => { + // 后序遍历函数 + const postOrder = node => { + // 递归出口 + if (!node) return [0, 0]; + // 遍历左子树 + const left = postOrder(node.left); + // 遍历右子树 + const right = postOrder(node.right); + // 不偷当前节点，左右子节点都可以偷或不偷，取最大值 + const DoNot = Math.max(left[0], left[1]) + Math.max(right[0], right[1]); + // 偷当前节点，左右子节点只能不偷 + const Do = node.val + left[0] + right[0]; + // [不偷，偷] + return [DoNot, Do]; + }; + const res = postOrder(root); + // 返回最大值 + return Math.max(...res); +}; +``` From 78c1d0209fc1bfcba73d52294b4b11db09f27f8a Mon Sep 17 00:00:00 2001 From: Jack <965555169@qq.com> Date: Thu, 5 Aug 2021 12:31:25 +0800 Subject: [PATCH 52/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A0121.=E4=B9=B0=E5=8D=96?= =?UTF-8?q?=E8=82=A1=E7=A5=A8=E7=9A=84=E6=9C=80=E4=BD=B3=E6=97=B6=E6=9C=BA?= =?UTF-8?q?-JavaScript=E7=89=88?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0121.买卖股票的最佳时机.md | 20 ++++++++++++++++++++ 1 file changed, 20 insertions(+) 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 259fff34..8ad6c8c6 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 @@ -313,6 +313,26 @@ func max(a,b int)int { } ``` +JavaScript： + +```javascript +const maxProfit = prices => { + const len = prices.length; + // 创建dp数组 + const dp = new Array(len).fill([0, 0]); + // dp数组初始化 + dp[0] = [-prices[0], 0]; + for (let i = 1; i < len; i++) { + // 更新dp[i] + dp[i] = [ + Math.max(dp[i - 1][0], -prices[i]), + Math.max(dp[i - 1][1], prices[i] + dp[i - 1][0]), + ]; + } + return dp[len - 1][1]; +}; +``` + From f351b222a5912d1972504b77c29b966e3be12e74 Mon Sep 17 00:00:00 2001 From: =?UTF-8?q?=E4=BD=99=E6=9D=9C=E6=9E=97?= Date: Thu, 5 Aug 2021 21:48:56 +0800 Subject: [PATCH 53/53] =?UTF-8?q?=E6=B7=BB=E5=8A=A00704.=E4=BA=8C=E5=88=86?= =?UTF-8?q?=E6=9F=A5=E6=89=BE=20Swift=E7=89=88=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0704.二分查找.md | 54 +++++++++++++++++++++++++++++++++++ 1 file changed, 54 insertions(+) diff --git a/problems/0704.二分查找.md b/problems/0704.二分查找.md index ed0e2eed..1bbb4b4f 100644 --- a/problems/0704.二分查找.md +++ b/problems/0704.二分查找.md @@ -352,6 +352,60 @@ def search(nums, target) end ``` +**Swift:** + +```swift +// （版本一）左闭右闭区间 +func search(nums: [Int], target: Int) -> Int { + // 1. 先定义区间。这里的区间是[left, right] + var left = 0 + var right = nums.count - 1 + + while left <= right {// 因为taeget是在[left, right]中，包括两个边界值，所以这里的left == right是有意义的 + // 2. 计算区间中间的下标（如果left、right都比较大的情况下，left + right就有可能会溢出） + // let middle = (left + right) / 2 + // 防溢出： + let middle = left + (right - left) / 2 + + // 3. 判断 + if target < nums[middle] { + // 当目标在区间左侧，就需要更新右边的边界值，新区间为[left, middle - 1] + right = middle - 1 + } else if target > nums[middle] { + // 当目标在区间右侧，就需要更新左边的边界值，新区间为[middle + 1, right] + left = middle + 1 + } else { + // 当目标就是在中间，则返回中间值的下标 + return middle + } + } + + // 如果找不到目标，则返回-1 + return -1 +} + +// （版本二）左闭右开区间 +func search(nums: [Int], target: Int) -> Int { + var left = 0 + var right = nums.count + + while left < right { + let middle = left + ((right - left) >> 1) + + if target < nums[middle] { + right = middle + } else if target > nums[middle] { + left = middle + 1 + } else { + return middle + } + } + + return -1 +} + +``` + -----------------------