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
退休钓鱼选手13 分钟前
[CommonAPI + vsomeip]通信 客户端 5
c++·人工智能·自动驾驶
what丶k32 分钟前
深度解析:以Kafka为例,消息队列消费幂等性的实现方案与生产实践
java·数据结构·kafka
星火开发设计36 分钟前
C++ 输入输出流:cin 与 cout 的基础用法
java·开发语言·c++·学习·算法·编程·知识
玖釉-39 分钟前
探索连续细节层次(Continuous LOD):深入解析 NVIDIA 的 nv_cluster_lod_builder
c++·windows·图形渲染
We་ct1 小时前
LeetCode 289. 生命游戏:题解+优化,从基础到原地最优
前端·算法·leetcode·矩阵·typescript
自己的九又四分之三站台1 小时前
9:MemNet记忆层使用,实现大模型对话上下文记忆
人工智能·算法·机器学习
sayang_shao1 小时前
C++ ONNX Runtime 与 Python Ultralytics 库实现 YOLOv8 模型检测的区别
c++·python·yolo
LXS_3571 小时前
STL - 函数对象
开发语言·c++·算法
aini_lovee1 小时前
基于粒子群算法(PSO)优化BP神经网络权值与阈值的实现
神经网络·算法
专注于ai算法的踩坑小达人1 小时前
C++变量全面总结
c++·qt