2019-05-02 15:59:01 +08:00
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# LeetCode 第 349 号问题:两个数组的交集
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> 本文首发于公众号「五分钟学算法」,是[图解 LeetCode ](<https://github.com/MisterBooo/LeetCodeAnimation>)系列文章之一。
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>
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> 个人网站:[https://www.cxyxiaowu.com](https://www.cxyxiaowu.com)
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题目来源于 LeetCode 上第 349 号问题:两个数组的交集。题目难度为 Easy,目前通过率为 62.3% 。
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### 题目描述
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给定两个数组,编写一个函数来计算它们的交集。
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**示例 1:**
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```
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输入: nums1 = [1,2,2,1], nums2 = [2,2]
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输出: [2]
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```
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**示例 2:**
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```
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输入: nums1 = [4,9,5], nums2 = [9,4,9,8,4]
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输出: [9,4]
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```
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**说明:**
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- 输出结果中的每个元素一定是唯一的。
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- 我们可以不考虑输出结果的顺序。
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### 题目解析
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容器类 [set](https://zh.cppreference.com/w/cpp/container/set) 的使用。
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- 遍历 num1,通过 set 容器 record 存储 num1 的元素
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- 遍历 num2 ,在 record 中查找是否有相同的元素,如果有,用 set 容器 resultSet 进行存储
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- 将 resultSet 转换为 vector 类型
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### 动画描述
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2019-11-14 11:00:28 +08:00
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![](https://blog-1257126549.cos.ap-guangzhou.myqcloud.com/blog/xfx1k.gif)
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2019-05-02 15:59:01 +08:00
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### 代码实现
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```
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class Solution {
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public:
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vector<int> intersection(vector<int>& nums1, vector<int>& nums2) {
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set<int> record;
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for(int i = 0; i < nums1.size(); i ++){
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record.insert(nums1[i]);
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}
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set<int> resultSet;
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for(int i = 0; i < nums2.size();i++){
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if(record.find(nums2[i]) != record.end()){
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resultSet.insert(nums2[i]);
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}
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}
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vector<int> resultVector;
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for(set<int>::iterator iter=resultSet.begin(); iter != resultSet.end();iter++){
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resultVector.push_back(*iter);
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}
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return resultVector;
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}
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};
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```
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2019-11-14 11:00:28 +08:00
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![](https://blog-1257126549.cos.ap-guangzhou.myqcloud.com/blog/y7jcl.png)
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