350. Intersection of Two Arrays IIEasy Topics Companies Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may return the result in any order. Example 1: Input: nums1 ...
publicstaticint[] intersect(int[] nums1,int[] nums2) { List<Integer> reslut =newArrayList<>(); Arrays.sort(nums1);//先进行数组排序Arrays.sort(nums2);intposition = 0;//记录位置for(inti = 0; i < nums2.length; i++) {if(position == nums1.length) {//终止条件:如果位置已经是最后...
leetcode 350. Intersection of Two Arrays II Given two arrays, write a function to compute their intersection. Example: Givennums1 =[1, 2, 2, 1],nums2 =[2, 2], return[2, 2]. Note: Each element in the result should appear as many times as it shows in both arrays. The result ca...
Can you solve this real interview question? Intersection of Two Arrays II - Given two integer arrays nums1 and nums2, return an array of their intersection. Each element in the result must appear as many times as it shows in both arrays and you may retur
LeetCode-Intersection of Two Arrays II Description: Given two arrays, write a function to compute their intersection. Example 1: Input: nums1 = [1,2,2,1], nums2 = [2,2] Output: [2,2] Example 2: Input: nums1 = [4,9,5], nums2 = [9,4,9,8,4]...
思路解析 解决方案一：这是Intersection of Two Arrays的加强版，同样使用HashMap。需要考虑交集中的重复次数。代码如下，时间复杂度为O(m+n）。解决方案二：思路为先对两个数组进行排序，然后使用两个指针分别指向nums1[]和nums2[]的开头，遇到相同的数字就加到ArrayList里，两个指针都增加，不相同则...
Intersection of Two Arrays II 编程算法 这个问题是之前一个问题的简单变形:传送门:LeetCode笔记:349. Intersection of Two Arrays。这个题目的变化在于他不要求每个数字只出现一次了,而是要求交叉了几次就保留几次,这其实更简单,还是之前题目的解法,把保障数字唯一性的代码去掉就好了,速度依然很快,3ms。 对于它提...
Given two arrays, write a function to compute their intersection. Example: Given nums1 = [1, 2, 2, 1], nums2 = [2, 2], return [2]. Note: 1、Each element in the result must be unique. 2、The result can be in any order. 要完成的函数: vector<int> intersection(vector<int>& ...
[i])){map.put(nums1[i],1);}else{map.put(nums1[i],map.get(nums1[i])+1);}}for(inti=0;i<length2;i++){if(map.containsKey(nums2[i])&&map.get(nums2[i])!=0){map.put(nums2[i],map.get(nums2[i])-1);ret[index++]=nums2[i];}}returnArrays.copyOfRange(ret,0,index);...
Problem link:https://leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/674/Solutions: 1. Sort and merge:We firstly sort the two arrays and use two pointers to compare the ele…