此博客停止维护和更新，已转移https://github.com/ilvseyinfu/blog

博主本来英文就不好...本来规则就不是很懂..先刷两道题试试水..在知乎上看说从ac率高往低刷..结果第一道题就参考了别人的..尴尬import java.util.*;public class Codec{ Map m =new HashMap(); public String encode(String longUrl){ String shortUrl = "http:/

/** * @param {string[]} ops * @return {number} */ var calPoints = function(ops) { var arr = []; for(var i =0 ;i<ops.length ;i++){ if(ops[i] == 'D'){ var leng

最近也是实在不知道该如何提升js编程水平，就用JavaScript刷LeetCode吧。发现自己有一个习惯，晚上吃完宵夜后，必须写代码，是要靠写代码助消化么？！ var findWords = function(words) { var arr2 =[]; for(var i=0 ;i<words.length ;i++){ var str = word

题意：求每一层的数据平均值，放到Vector里思路：层序遍历，没啥难的/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), le

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */clas

反转二叉树，没仔细看题，层序遍历的思想，写完WA了，一看错误数据，尴尬的反转错了 TreeNode* invertTree(TreeNode* root) { queue q; TreeNode* newroot = root; q.push(root); while(!q.empty()){

合并两颗二叉树，已经发现解决二叉树问题的利器就是递归～/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL),

驾照GET，最近一直在准备笔试面试，很少有时间去看科四，回家高铁上才开始看，第二天考100分过哈哈，希望可以找到一份满意的工作JS实现先说两个函数;arr.reverse()  反转数组，反转后还是数组str.split('') 将字符串转换为数组arr.join('') 将数组元素转化为字符串 var arr = ['a','b','c','d'];var str =

Given an array of integers where 1 ≤ a[i] ≤ n (n = size of array), some elements appear twice and others appear once.Find all the elements of [1, n] inclusive that do not appear in this array.

O(n)的时间复杂的没有想出来，看了Discuss.int singleNumber(int* nums, int numsSize) { int num = 0; for(int i=0 ;i<numsSize; i++){ num ^= nums[i]; } return num;}

/// 实现链表反转的两种方法/// 1. 创建新链表struct ListNode* reverseList(struct ListNode* head) { if(head == NULL) return NULL; struct ListNode* nlist = NULL; for(struct ListNode* p = head;p;){

237. Delete Node in a Linked ListWrite a function to delete a node (except the tail) in a singly linked list, given only access to that node.Supposed the linked list is 1 -> 2 -> 3 -> 4 and

日常刷leetcode，一道简单题，脑子不是很清醒啊int findMaxConsecutiveOnes(int* nums, int numsSize) { int count = 0 ; int result = 0; for(int i=0 ;i<numsSize ; i++){ if(nums[i]){ count++;

递归求解，每次找出最的的元素作为树根，思路有点类似快排class Solution { /** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x)

日常刷leetcode.../** * @param {number[]} nums * @return {number[]} */var findDuplicates = function(nums) { var arr = []; for(var i=0 ;i<nums.length; i++){ var index = Math.abs(nums[

勉强算一道搜索吧。。class Solution {public: int islandPerimeter(vector>& grid) { int row = grid.size(); int col = grid[0].size(); int sum = 0; for(int i=0 ;i<row ; i++){