因为使用Arrays类的sort方法,所以时间复杂度是O(n log(n)),空间复杂度是O(n)。 publicint[]intersection3(int[] nums1,int[] nums2){ Set<Integer>set=newHashSet<>(); Arrays.sort(nums1); Arrays.sort(nums2);inti =0;intj =0;while(i < nums1.length && j < nums2.length) {if(nums1[i...
先将两数组排序,然后使用双指针,依次判断两数组中的元素是否相等,如果某个元素大于或小于另外一个元素,则将指针向后移动,如果相等,则将元素放入ArrayList中,然后将ArrayList中的元素迭代放入数组,最后返回。 因为使用Arrays类的sort方法,所以时间复杂度是O(n log(n)),空间复杂度是O(n)。 publicint[]intersect(int...
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>& ...
publicclassSolution{publicint[]intersect(int[]nums1,int[]nums2){List<Integer>list=newArrayList<Integer>();intlength1=nums1.length;intlength2=nums2.length;int[]ret=newint[Math.min(length1,length2)];intindex=0;HashMap<Integer,Integer>map=newHashMap<Integer,Integer>();for(inti=0;i<length1...
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 = [1,2,2,1], nums2 = [2,2] Output: [2,2] ...
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]...
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
1// 349. Intersection of Two Arrays 2// https://leetcode.com/problems/intersection-of-two-arrays/description/ 3// 时间复杂度: O(nlogn) 4// 空间复杂度: O(n) 5class Solution { 6public: 7 vector<int> intersection(vector<int>& nums1, vector<int>& nums2) { 8 9 set<int> record(...
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...
简介:LeetCode之Intersection of Two Arrays 1、题目 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. ...