LeetCode75——Day20

文章目录

一、题目

2215. Find the Difference of Two Arrays

Given two 0-indexed integer arrays nums1 and nums2, return a list answer of size 2 where:

answer[0] is a list of all distinct integers in nums1 which are not present in nums2.

answer[1] is a list of all distinct integers in nums2 which are not present in nums1.

Note that the integers in the lists may be returned in any order.

Example 1:

Input: nums1 = [1,2,3], nums2 = [2,4,6]

Output: [[1,3],[4,6]]

Explanation:

For nums1, nums1[1] = 2 is present at index 0 of nums2, whereas nums1[0] = 1 and nums1[2] = 3 are not present in nums2. Therefore, answer[0] = [1,3].

For nums2, nums2[0] = 2 is present at index 1 of nums1, whereas nums2[1] = 4 and nums2[2] = 6 are not present in nums2. Therefore, answer[1] = [4,6].

Example 2:

Input: nums1 = [1,2,3,3], nums2 = [1,1,2,2]

Output: [[3],[]]

Explanation:

For nums1, nums1[2] and nums1[3] are not present in nums2. Since nums1[2] == nums1[3], their value is only included once and answer[0] = [3].

Every integer in nums2 is present in nums1. Therefore, answer[1] = [].

Constraints:

1 <= nums1.length, nums2.length <= 1000

-1000 <= nums1[i], nums2[i] <= 1000

二、题解

cpp 复制代码
class Solution {
public:
    vector<vector<int>> findDifference(vector<int>& nums1, vector<int>& nums2) {
        int n1 = nums1.size();
        int n2 = nums2.size();
        vector<int> res1;
        vector<int> res2;
        unordered_map<int,int> map1;
        unordered_map<int,int> map2;
        for(int i = 0;i < n1;i++) map1[nums1[i]]++;
        for(int i = 0;i < n2;i++) map2[nums2[i]]++;
        for(int i = 0;i < n1;i++){
            if(map2[nums1[i]] == 0){
                res1.push_back(nums1[i]);
                map2[nums1[i]]++;
            }
        }
        for(int i = 0;i < n2;i++){
            if(map1[nums2[i]] == 0){
                res2.push_back(nums2[i]);
                map1[nums2[i]]++;
            }
        }
        vector<vector<int>> res;
        res.push_back(res1);
        res.push_back(res2);
        return res;
    }
};
相关推荐
颜酱26 分钟前
一步步实现字符串计算器:从「转整数」到「带括号与优化」
javascript·后端·算法
不想写代码的星星1 小时前
std::function 详解:用法、原理与现代 C++ 最佳实践
c++
CoovallyAIHub19 小时前
语音AI Agent编排框架!Pipecat斩获10K+ Star,60+集成开箱即用,亚秒级对话延迟接近真人反应速度!
深度学习·算法·计算机视觉
NineData20 小时前
数据库管理工具NineData,一年进化成为数万+开发者的首选数据库工具?
运维·数据结构·数据库
木心月转码ing1 天前
Hot100-Day14-T33搜索旋转排序数组
算法
会员源码网1 天前
内存泄漏(如未关闭流、缓存无限增长)
算法
颜酱1 天前
从0到1实现LFU缓存:思路拆解+代码落地
javascript·后端·算法
颜酱1 天前
从0到1实现LRU缓存:思路拆解+代码落地
javascript·后端·算法
CoovallyAIHub2 天前
Moonshine:比 Whisper 快 100 倍的端侧语音识别神器,Star 6.6K!
深度学习·算法·计算机视觉