(for the scope of this problem, consider only the number with equal digits) Then the lunar sum of a and b will be − 879 We are required to write a JavaScript function that takes in two numbers and returns their lunar sum. Advertisement - This is a modal window. No compatible source...
这里面还可以做一些小的优化,比如把length拿出来,重复使用的nums[i]也抽取出来,遍历的顺序反过来等,最后大概弄成这个样子: vartwoSum2 =function(nums, target) {varhash ={};vari;vartmp;for(i = nums.length; i--; ) { tmp=nums[i];if(typeofhash[tmp] !== "undefined") {return[i, hash[tmp]...
/** * @param {number[]} nums * @param {number} target * @return {number[]} */ const twoSum = function (nums, target) { const map = new Map(); const len = nums.length; //数组长度 for (let i = 0; i < len; i++) { //遍历查找数组nums let diff = target - nums[i]; ...
* @param {number[]} nums * @param {number} target * @return {number[]} */ var twoSum = function (nums, target) { function isEqualToTarget(num1, num2) { if (num1 + num2 === target) { return true } else { return false } } function findTargetItem(arr, oldArr) { for (let...
每天一算:Two Sum II leetcode上167号问题:Two Sum II 给定一个已按照升序排列 的有序数组,找到两个数使得它们相加之和等于目标数。 函数应该返回这两个下标值 index1 和 index2,其中 index1 必须小于 index2。 说明: 返回的下标值(index1 和 index2)不是从零开始的。
Inspired by Chris' discussion on the topic in episode 77 of Good Job BrainInputs First Number Second NumberResults Digital Root of 1st Number Digital Root of 2nd Number Sum of the two numbers Sum of the digital roots Root of the sum of the digital roots Digital root of the sum...
思路一的 indexOf() 时间复杂度是 O(n),故思路一时间复杂度为 O(n*n),而 JavaScript 的对象属性查找是哈希查找,时间复杂的是 O(1),思路二的时间复杂度为 O(n)。下面是代码:/** * @param {number[]} nums * @param {number} target * @return {number[]} */ var twoSum = function(...
While it may not be as concise, it offers clarity in terms of what each part of the code is doing. function sumArray(arr) { let sum = 0; arr.forEach((number) => { sum += number; }); return sum; } const numbers = [1, 2, 3, 4, 5]; console.log(sumArray(numbers)); ...
[i]; let remainIndex = cost.indexOf(remain,i+1); if(remainIndex!=-1){ let first = cpyarr.indexOf(cost[i]); let last = cpyarr.indexOf(cost[remainIndex]); if(last === first){ last = cpyarr.indexOf(cost[remainIndex],first)+2; first++; }else{ last++; first++; }; ...
Given two integersnandk, return all possible combinations ofknumbers out of 1 ...n. Example: 代码语言:javascript 代码运行次数:0 运行 AI代码解释 Input:n=4,k=2Output:[[2,4],[3,4],[2,3],[1,2],[1,3],[1,4],] 题目大意:在1 ...n中任意选取k个数,将所有的情况罗列出来。