// Function to find the union of two arraysconstunion=(arra1,arra2)=>{// Check if either of the arrays is null, return undefined if trueif((arra1==null)||(arra2==null))returnvoid0;// Initialize an empty object to store unique elements from both arraysconstobj={};// Iterate throu...
使用Underscore.js 或 Lo-Dash,您可以: _.union([1,2,3], [101,2,1,10], [2,1]); => [1,2,3,101,10] http://underscorejs.org/#union http://lodash.com/docs#union
- madhu Goud1 使用lodash union https://lodash.com/docs/4.17.10#union 或 https://lodash.com/docs/4.17.11#uniqWith。 - CornelC 你可以简单地循环遍历元素较少的数组并与另一个数组进行比较。将不存在的对象推送到同一数组中,以避免使用第三个变量。 - super cool...
Write a JavaScript program to compute the union of two arrays. Sample Data : console.log(union([1, 2, 3], [100, 2, 1, 10])); [1, 2, 3, 10, 100] Click me to see the solution 23. Difference Between Arrays Write a JavaScript function to find the difference between two arrays. ...
如何在JavaScript中合并两个数组并重复删除项目今天 2020.10.15 我在 Chrome v86、Safari v13.1.2 和 Firefox v81 上针对所选解决方案在 MacOs HighSierra 10.13.6 上执行测试。
http://lodash.com/docs#union http://lodash.com/docs#union #3楼 Just throwing in my two cents. 只需投入我的两分钱。 function mergeStringArrays(a, b){ var hash = {}; var ret = []; for(var i=0; i < a.length; i++){ var e = a[i]; if (!hash[e]){ hash[e] = true...
https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/ 描述 给定两个数组,编写一个函数来计算它们的交集。 示例 1: 示例 2: 说明: 输出结果中每个元素出现的次数,应与元素在两个数组中出现的次数一致。 我们可以不考虑输出结果的顺序。 进阶: 如果给定的数组已经排好序呢?你将如何优化你的算法?
union(array): Returns the union of two arrays. intersect(array): Returns the intersection of two arrays. difference(array): Returns the difference of two arrays. zip(array): Zips two arrays together. flatten(): Flattens a nested array. ...
union Returns every element that exists in any of the two arrays once. Create aSetwith all values ofaandband convert to an array. const union = (a, b) => Array.from(new Set([...a, ...b])); 返回两个数组的并集(像集合的并集一样,不包含重复元素)。
实现并集(Union)、交集(Intersect)和差集(Difference) let a = new Set([1, 2, 3]); let b = new Set([4, 3, 2, 1]); // 并集 let union = new Set([...a, ...b]); // Set {1, 2, 3, 4} // 交集 let intersect = new Set([...a].filter(x => b.has(x))); ...