1、这道题给定两个vector,要求返回两个vector的交集,比如nums1=[1,2,2,1],nums2=[2,2],返回的交集是[2,2],其中有多少个相同的元素就返回多少个。返回的交集不讲究顺序。 2、这道题看完题意,熟悉leetcode的同学应该会马上想到先排序,排序之后的两个vector来比较,时间复杂度会下降很多。 如果不排序,那就...
思路:偷懒一下,用Intersection of Two Arrays 的算法改动一下,看了下讨论,基本也是用hash,欢迎交流不同解法 public int[] intersect(int[] nums1, int[] nums2) { int len1 = nums1.length; int len2 = nums2.length; if(len1==0 || len2==0){ return new int[0]; } HashMap<Integer,Integer...
What if elements of nums2 are stored on disk, and the memory is limited such that you cannot load all elements into the memory at once? 思路: Solution 1: 这道题就是Intersection of Two Arrays的加强版。同样用HashMap来做。需要考虑在intersection中重复的次数。 代码如下: public class Solution ...
Leetcode 350:Intersection of Two Arrays II Given two arrays, write a function to compute their intersection. 说人话: 求两个数组的交集。 举例1: 举例2: [法1] Map 记录频次 思路 因为是求交集,所以元素出现的次数是需要进行考虑了。这里笔者考虑到用 Map 这个集合框架来解决这个问题,主要是基于 key-v...
350. 两个数组的交集 II Intersection of Two Arrays II难度:Easy| 简单 相关知识点:排序 哈希表 双指针 二分查找题目链接:https://leetcode-cn.com/problems/intersection-of-two-arrays-ii/官方题解:https://leetcode-cn.com/problems/intersection-of-
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] 350. Intersection of Two Arrays II,Giventwoarrays,writeafunctiontocomputetheirintersection.Example1:Input:nums1=[1,2,2,1],nums2=[2,2]Output:[2,2]Example2:Input:n
Each element in the result should appear as many times as it shows in both arrays. The result can be in any order. 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: 输入: nums1 = [1,2,2,1], nums2 = [2,2] 输出: [2,2] ...
https://leetcode.com/explore/interview/card/top-interview-questions-easy/92/array/674/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 elements in the two arrays...
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...