diff --git a/problems/0142.环形链表II.md b/problems/0142.环形链表II.md index 7556b854..b3a53c07 100644 --- a/problems/0142.环形链表II.md +++ b/problems/0142.环形链表II.md @@ -234,6 +234,29 @@ class Solution: ``` Go： +```func detectCycle(head *ListNode) *ListNode { + if head ==nil{ + return head + } + slow:=head + fast:=head.Next + + for fast!=nil&&fast.Next!=nil{ + if fast==slow{ + slow=head + fast=fast.Next + for fast!=slow { + fast=fast.Next + slow=slow.Next + } + return slow + } + fast=fast.Next.Next + slow=slow.Next + } + return nil +} +``` ----------------------- diff --git a/problems/0216.组合总和III.md b/problems/0216.组合总和III.md index 21230e0f..bcb443e3 100644 --- a/problems/0216.组合总和III.md +++ b/problems/0216.组合总和III.md @@ -94,7 +94,7 @@ void backtracking(int targetSum, int k, int sum, int startIndex) 所以 终止代码如下： -``` +```C++ if (path.size() == k) { if (sum == targetSum) result.push_back(path); return; // 如果path.size() == k 但sum != targetSum 直接返回 @@ -112,7 +112,7 @@ if (path.size() == k) { 代码如下： -``` +```C++ for (int i = startIndex; i <= 9; i++) { sum += i; path.push_back(i); @@ -126,7 +126,7 @@ for (int i = startIndex; i <= 9; i++) { 参照[关于回溯算法，你该了解这些！](https://mp.weixin.qq.com/s/gjSgJbNbd1eAA5WkA-HeWw)中的模板，不难写出如下C++代码： -``` +```C++ class Solution { private: vector> result; // 存放结果集 @@ -273,4 +273,4 @@ Go： * 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) -
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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 29bcdf41..1180225a 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 @@ -324,8 +324,31 @@ class Solution { ``` Python： +```Python +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def binaryTreePaths(self, root: TreeNode) -> List[str]: + path=[] + res=[] + def backtrace(root, path): + if not root:return + path.append(root.val) + if (not root.left)and (not root.right): + res.append(path[:]) + ways=[] + if root.left:ways.append(root.left) + if root.right:ways.append(root.right) + for way in ways: + backtrace(way,path) + path.pop() + backtrace(root,path) + return ["->".join(list(map(str,i))) for i in res] - +``` Go： diff --git a/problems/0383.赎金信.md b/problems/0383.赎金信.md index 742bfdc3..2f7ae4d8 100644 --- a/problems/0383.赎金信.md +++ b/problems/0383.赎金信.md @@ -26,7 +26,7 @@ canConstruct("a", "b") -> false canConstruct("aa", "ab") -> false canConstruct("aa", "aab") -> true -# 思路 +## 思路 这道题目和[242.有效的字母异位词](https://mp.weixin.qq.com/s/vM6OszkM6L1Mx2Ralm9Dig)很像，[242.有效的字母异位词](https://mp.weixin.qq.com/s/vM6OszkM6L1Mx2Ralm9Dig)相当于求 字符串a 和 字符串b 是否可以相互组成 ，而这道题目是求 字符串a能否组成字符串b，而不用管字符串b 能不能组成字符串a。 @@ -36,7 +36,7 @@ canConstruct("aa", "aab") -> true * 第二点 “你可以假设两个字符串均只含有小写字母。” *说明只有小写字母*，这一点很重要 -# 暴力解法 +## 暴力解法 那么第一个思路其实就是暴力枚举了，两层for循环，不断去寻找，代码如下： @@ -67,7 +67,7 @@ public: 这里时间复杂度是比较高的，而且里面还有一个字符串删除也就是erase的操作，也是费时的，当然这段代码也可以过这道题。 -# 哈希解法 +## 哈希解法 因为题目所只有小写字母，那可以采用空间换取时间的哈希策略， 用一个长度为26的数组还记录magazine里字母出现的次数。 @@ -105,8 +105,6 @@ public: - - ## 其他语言版本 diff --git a/problems/0404.左叶子之和.md b/problems/0404.左叶子之和.md index 0290dccd..5f303196 100644 --- a/problems/0404.左叶子之和.md +++ b/problems/0404.左叶子之和.md @@ -205,8 +205,25 @@ 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 +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 +``` Go： diff --git a/problems/0406.根据身高重建队列.md b/problems/0406.根据身高重建队列.md index 2ef0a2fd..c404ec8d 100644 --- a/problems/0406.根据身高重建队列.md +++ b/problems/0406.根据身高重建队列.md @@ -211,7 +211,18 @@ class Solution { ``` Python： - +```python +class Solution: + def reconstructQueue(self, people: List[List[int]]) -> List[List[int]]: + people.sort(key=lambda x: (x[0], -x[1]), reverse=True) + que = [] + for p in people: + if p[1] > len(que): + que.append(p) + else: + que.insert(p[1], p) + return que +``` Go： @@ -222,4 +233,4 @@ Go： * 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) -
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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 ac21dc68..44096a29 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 +274,28 @@ class Solution { Python： - - +```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] +``` Go： @@ -285,4 +305,4 @@ Go： * 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) -
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diff --git a/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md b/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md index 3926643d..a239e5d8 100644 --- a/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md +++ b/problems/0714.买卖股票的最佳时机含手续费（动态规划）.md @@ -95,7 +95,48 @@ public: Java： +```java +/** + * 卖出时支付手续费 + * @param prices + * @param fee + * @return + */ +public int maxProfit(int[] prices, int fee) { + int len = prices.length; + // 0 : 持股（买入） + // 1 : 不持股（售出） + // dp 定义第i天持股/不持股 所得最多现金 + int[][] dp = new int[len][2]; + dp[0][0] = -prices[0]; + for (int i = 1; i < len; i++) { + dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] - prices[i]); + dp[i][1] = Math.max(dp[i - 1][0] + prices[i] - fee, dp[i - 1][1]); + } + return Math.max(dp[len - 1][0], dp[len - 1][1]); +} +/** + * 买入时支付手续费 + * @param prices + * @param fee + * @return + */ +public int maxProfit(int[] prices, int fee) { + int len = prices.length; + // 0 : 持股（买入） + // 1 : 不持股（售出） + // dp 定义第i天持股/不持股 所得最多现金 + int[][] dp = new int[len][2]; + // 考虑买入的时候就支付手续费 + dp[0][0] = -prices[0] - fee; + for (int i = 1; i < len; i++) { + dp[i][0] = Math.max(dp[i - 1][0], dp[i - 1][1] - prices[i] - fee); + dp[i][1] = Math.max(dp[i - 1][0] + prices[i], dp[i - 1][1]); + } + return Math.max(dp[len - 1][0], dp[len - 1][1]); +} +``` Python： diff --git a/problems/0860.柠檬水找零.md b/problems/0860.柠檬水找零.md index bf8a3776..052bcdb1 100644 --- a/problems/0860.柠檬水找零.md +++ b/problems/0860.柠檬水找零.md @@ -156,7 +156,30 @@ class Solution { ``` Python： +```python +class Solution: + def lemonadeChange(self, bills: List[int]) -> bool: + five, ten, twenty = 0, 0, 0 + for bill in bills: + if bill == 5: + five += 1 + elif bill == 10: + if five < 1: return False + five -= 1 + ten += 1 + else: + if ten > 0 and five > 0: + ten -= 1 + five -= 1 + twenty += 1 + elif five > 2: + five -= 3 + twenty += 1 + else: + return False + return True +``` Go： @@ -167,4 +190,4 @@ Go： * 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) -
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diff --git a/problems/1143.最长公共子序列.md b/problems/1143.最长公共子序列.md index 19f4cc72..961924c8 100644 --- a/problems/1143.最长公共子序列.md +++ b/problems/1143.最长公共子序列.md @@ -166,6 +166,34 @@ class Solution: Go： +```Go +func longestCommonSubsequence(text1 string, text2 string) int { + t1 := len(text1) + t2 := len(text2) + dp:=make([][]int,t1+1) + for i:=range dp{ + dp[i]=make([]int,t2+1) + } + + for i := 1; i <= t1; i++ { + for j := 1; j <=t2; j++ { + if text1[i-1]==text2[j-1]{ + dp[i][j]=dp[i-1][j-1]+1 + }else{ + dp[i][j]=max(dp[i-1][j],dp[i][j-1]) + } + } + } + return dp[t1][t2] +} + +func max(a,b int)int { + if a>b{ + return a + } + return b +} +``` @@ -174,4 +202,4 @@ Go： * 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) -
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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 97045ed2..493438d7 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 @@ -92,7 +92,63 @@ public: Java： - +```Java +/** + * Definition for singly-linked list. + * public class ListNode { + * int val; + * ListNode next; + * ListNode(int x) { + * val = x; + * next = null; + * } + * } + */ +public class Solution { + public ListNode getIntersectionNode(ListNode headA, ListNode headB) { + ListNode curA = headA; + ListNode curB = headB; + int lenA = 0, lenB = 0; + while (curA != null) { // 求链表A的长度 + lenA++; + curA = curA.next; + } + while (curB != null) { // 求链表B的长度 + lenB++; + curB = curB.next; + } + curA = headA; + curB = headB; + // 让curA为最长链表的头，lenA为其长度 + if (lenB > lenA) { + //1. swap (lenA, lenB); + int tmpLen = lenA; + lenA = lenB; + lenB = tmpLen; + //2. swap (curA, curB); + ListNode tmpNode = curA; + curA = curB; + curB = tmpNode; + } + // 求长度差 + int gap = lenA - lenB; + // 让curA和curB在同一起点上（末尾位置对齐） + while (gap-- > 0) { + curA = curA.next; + } + // 遍历curA 和 curB，遇到相同则直接返回 + while (curA != null) { + if (curA == curB) { + return curA; + } + curA = curA.next; + curB = curB.next; + } + return null; + } + +} +``` Python：