Jason_312的博客

版本一class Solution {public: ListNode* deleteDuplicates(ListNode* head) { if(head==NULL)return head; ListNode * L=new ListNode(-1); L->next=head; while(head->next!=NULL){ ListNode* node=head->next;.

给定一个链表，两两交换其中相邻的节点，并返回交换后的链表。你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。示例 1：输入：head = [1,2,3,4]输出：[2,1,4,3]class Solution {public: ListNode* swapPairs(ListNode* head) { //递归 //结束条件 head==null||head->next==null // 本层递归 i

class Solution {public: void dfs(TreeNode* root,vector<int>& v){ if(root==nullptr)return ; dfs(root->left,v); v.push_back(root->val); dfs(root->right,v); } TreeNode* increasingBST(TreeNode* r.

class Solution {public: int maxDepth(TreeNode* root) { if(root==nullptr)return 0; queue<TreeNode*>q; q.push(root); int ans=0; while(!q.empty()){ int t=q.size();//每层的节点数 while(t--){ .

第一次提交class Solution {public: int uniquePathsWithObstacles(vector<vector<int>>& obstacleGrid) { int dp[100][100]; int m=obstacleGrid.size(); int n=obstacleGrid[0].size(); if(obstacleGrid[0][0]==

You are given an array prices where prices[i] is the price of a given stock on the ith day.You want to maximize your profit by choosing a single day to buy one stock and choosing a different day in the future to sell that stock.Return the maximum profit

You are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed. All houses at this place are arranged in a circle. That means the first house is the neighbor of the last one. Meanwhile, adjacent houses

Given a triangle array, return the minimum path sum from top to bottom.For each step, you may move to an adjacent number of the row below. More formally, if you are on index i on the current row, you may move to either index i or index i + 1 on the next r

198. House RobberYou are a professional robber planning to rob houses along a street. Each house has a certain amount of money stashed, the only constraint stopping you from robbing each of them is that adjacent houses have security systems connected and