方法一:最普遍的做法 使用 ES5 语法来实现虽然会麻烦些,但兼容性最好,不用考虑浏览器 JavaScript 版本.也不用引入其他第三方库. 1,直接使用 filter.concat 来计算 var a = [1,2,3,4,5] var b = [2,4,6,8,10] //交集 var c = a.filter(function(v){ return b.indexOf(v) > -1 }) //差集 var d = a.filter(function(v){ return b.i
#!/usr/bin/perl use strict; ######################################## 用grep 和map 获取两个列表的交集并集.补集####################################### my @a=("a","b","c","d","e");my @b=("b","g","f&
问题分析: 用两个指针分别遍历即可. 问题求解: public class Solution { /** * @param nums1 an integer array * @param nums2 an integer array * @return an integer array */ public int[] intersection(int[] nums1, int[] nums2) { List<Integer> list = new ArrayList<Integer>
问题分析: 用两个指针分别遍历即可. 问题求解: public class Solution { /** * @param nums1 an integer array * @param nums2 an integer array * @return an integer array */ public int[] intersection(int[] nums1, int[] nums2) { List<Integer> list = new ArrayList<Integer>
1.考虑不重复元素,重复元素不添加 import java.awt.List; import java.util.ArrayList; import java.util.TreeSet; public class Solution { public static int[] intersection(int[] nums1,int[] nums2){ TreeSet<Integer> set =new TreeSet<>(); for(int num : nums1)//把不重复的
Given two arrays, write a function to compute their intersection. Example:Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2]. Note: Each element in the result must be unique. The result can be in any order. 这道题让我们找两个数组相同的部分,难度不算大,我们可以用个set把nums1都
Problem: Given two arrays, write a function to compute their intersection. 中文:已知两个数组,写一个函数来计算它们的交集 Example: Given nums1 = [1, 2, 2, 1], nums2 = [2, 2],return [2, 2]. 已知nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2, 2]. Note: Each element in the res
Given two arrays, write a function to compute their intersection. Example 1: Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2] Example 2: Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4] Output: [9,4] Note: Each element in the result must be unique. The