april_4的博客

var combinationSum = function(candidates, target) { const len = candidates.length; const ret = []; const dfs = (start, buf=[],target) => { for(let index = start;index < len;index++){ buf.push(candidates[index]); .

给定一个非空的整数数组，返回其中出现频率前k高的元素。示例 1:输入: nums = [1,1,1,2,2,3], k = 2输出: [1,2]示例 2:输入: nums = [1], k = 1输出: [1]提示：你可以假设给定的k总是合理的，且 1 ≤ k ≤ 数组中不相同的元素的个数。你的算法的时间复杂度必须优于 O(n log n) ,n是数组的大小。题目数据保证答案唯一，换句话说，数组中前 k 个高频元素的集合是唯一的。你可以按任意顺序返回答案...

由于不可以在相邻的房屋闯入，所以在当前位置 n 房屋可盗窃的最大值，要么就是 n-1 房屋可盗窃的最大值，要么就是 n-2 房屋可盗窃的最大值加上当前房屋的值，二者之间取最大值var rob = function(nums) { const len = nums.length; if(len === 0) return 0; let dp=[];...

var maxProfit = function(prices) { let min = prices[0]; let max =0; let len = prices.length; for (let index=1;index<len;index++){ if(prices[index]< min){ min...

var maxSubArray = function(nums) { let ans = nums[0]; let len = nums.length; let sum = 0; for(let index = 0;index < len;index++){ if(sum > 0){ sum +=nums[ind...

var groupAnagrams = function(strs) { let hashmap = new Map() for(let index = 0;index < strs.length;index++){ let str = strs[index].split('').sort().join(); if(hashmap.has(...

function divide(x,n){ if(n == 0) return 1; let subresult = divide(x,parseInt(n/2)) if(n&1 === 1){ return subresult*subresult*x; }else{ return subresult*subres...

var addStrings = function(num1, num2) { let str =''; let carry = 0; let i = num1.length - 1; let j = num2.length - 1; while(i>= 0 || j >=0){ let n1 = i>= 0?num1[i...

var combinationSum2 = function(candidates, target) { candidates.sort((a,b)=>{return b-a}) let len = candidates.length let min = candidates[len-1]; let res=[]; let path =[]; ...

var combinationSum = function (candidates, target) { candidates.sort((a,b)=>{return b-a}) let len = candidates.length let min = candidates[len-1]; let res=[]; let path =[]; ...

原题解来自https://leetcode-cn.com/problems/count-and-say/solution/jsdi-gui-jie-jue-by-wo-shi-gao-xiao-de/看了这个哥们的解法，觉得特别好，顺便复习了下正则var countAndSay = function(n) { if(n==1){ return n.toStrin...

var searchInsert = function(nums, target) { let left = 0; let right = nums.length-1; if(target > nums[right]) return nums.length; while(left < right){ let mid ...

function gcd(a, b) { return a % b === 0 ? b : gcd(b, a % b);}var hasGroupsSizeX = function(deck) { let len = deck.length; if(len <= 1 || len> 10000) return false; cons...

var longestCommonPrefix = function(strs) { let len = strs.length; let str = ''; if(len === 0){ return '' } if(len === 1){ return strs[0] } str = findPublicPrefix(strs[0],...

木桶原则只循环一次，算法关键点是理解体积 受制于最短的板var maxArea = function(height) { let l = 0; let r=height.length - 1; let maxarea = 0; while(l < r){ maxarea = Math.max(maxarea,(r-l)*Math.m...

var reverse = function(x) { let str = x.toString(); if(str.length == 0){ return 0; } let asc = str[0] let final = '' if(asc === '-' || asc === '+'){ st...

最近更新频率小了，感觉这样下去就荒废了，在做好本职工作的前提下，还是应该有所提升和进步，再次开始刷领扣吧~定个规矩，还是先自己想自己写，然后再看人家的做法，即使自己写的烂也好歹是自己的想法，今天做了整数转罗马数字，我是这么写的，有点繁琐，不过还算一遍对了/** * @param {number} num * @return {string} */function handleThou...

var myAtoi = function(str) { let desStr = str.trim(); if(desStr.length === 0 ){ return 0; } let asc = desStr[0]; if(asc != '+' && asc != '-' && !/^\d...

之前想用三重循环，或者用Hash的方法来解决这个问题，但是没办法去重后来还是参考了下别人的var threeSum = function(nums) { let num = nums.sort((a, b) => a - b); let res = [] if(num.length === 0) return res if(num...

和昨天的那道题目很像，理解了后这道题20分钟就出来了var threeSumClosest = function(nums, target) { nums.sort((a,b) => a-b); let size = nums.length; let keepminus = undefined; for(let i...

这个题目用到了reduce和map，还了解了下这种排列组合叫做笛卡尔卷积，以后这种组合的就可以多考虑用用reducevar letterCombinations = function(digits) { if(digits === '') return []; let hash = { '2':['a','b','c'], '3':[...

头发蒙，今天不知道失败了多少次才做出来。。。双指针又是var removeNthFromEnd = function(head, n) { let first = head; let count = 0; let second = head; if(head.next == null){ return null }...

var isValid = function (s) { var map = { "(": ")", "[": "]", "{": "}" } var leftArr = [] for (var ch of s){ if (ch in map) leftArr.push(ch); //为左括号时，顺序...

被这个程序的描述误导了有点，其实要求返回的是不重复的数组个数var removeDuplicates = function(nums) { let len = nums.length; let newLen = 0; for(let index =0; index <len;index++){ nums[newLen] = nums[index...

今天撸的第二道题，动脑的感觉不错，照例写了一大坨，看来要好好看看算法了思路就是取两个字符，看是否在字典中，如果有，就累计，并且跳过两个走下一轮，如果没有就查单个字符的字典累计，不过虽然麻烦，但是跑起来用时比下面写的经典的低100多毫秒，（用了slice，后面才想起来还有substr。。。）var romanToInt = function(s) { let nums = [100...