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# 0349. 两个数组的交集

### 题目地址(349. 两个数组的交集)

<https://leetcode-cn.com/problems/intersection-of-two-arrays/>

### 题目描述

```
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]
示例 2：

输入：nums1 = [4,9,5], nums2 = [9,4,9,8,4]
输出：[9,4]
 

说明：

输出结果中的每个元素一定是唯一的。
我们可以不考虑输出结果的顺序。

```

### 前置知识

* hashtable

### 公司

* 阿里
* 腾讯
* 百度
* 字节

### 思路

先遍历第一个数组，将其存到 hashtable 中，然后遍历第二个数组，如果在 hashtable 中存在就 push 到 ret，然后清空 hashtable，最后返回 ret 即可。

### 关键点解析

* 空间换时间

### 代码

代码支持：JS, Python

Javascript Code:

```js
/**
 * @param {number[]} nums1
 * @param {number[]} nums2
 * @return {number[]}
 */
var intersection = function (nums1, nums2) {
  const visited = {};
  const ret = [];
  for (let i = 0; i < nums1.length; i++) {
    const num = nums1[i];

    visited[num] = num;
  }

  for (let i = 0; i < nums2.length; i++) {
    const num = nums2[i];

    if (visited[num] !== undefined) {
      ret.push(num);
      visited[num] = undefined;
    }
  }

  return ret;
};
```

Python Code:

```python
class Solution:
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        visited, result = {}, []
        for num in nums1:
            visited[num] = num
        for num in nums2:
            if num in visited:
                result.append(num)
                visited.pop(num)
        return result

    # 另一种解法：利用 Python 中的集合进行计算
    def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]:
        return set(nums1) & set(nums2)
```

**复杂度分析**

* 时间复杂度：$O(N)$
* 空间复杂度：$O(N)$

更多题解可以访问我的 LeetCode 题解仓库：<https://github.com/azl397985856/leetcode> 。 目前已经 37K star 啦。

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