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349-intersection-of-two-arrays.md Separated Chinese solutions from English solutions.
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‎en/1-1000/349-intersection-of-two-arrays.md

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#349. Intersection of Two Arrays - Best Practices of LeetCode Solutions
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LeetCode link:[349. Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays),
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[349. 两个数组的交集](https://leetcode.cn/problems/intersection-of-two-arrays)
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LeetCode link:[349. Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays), difficulty:**Easy**.
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[中文题解](#中文题解)
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##LeetCode problem description
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##LeetCode description of "349. Intersection of Two Arrays"
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Given two integer arrays`nums1` and`nums2`, return_an array of their**intersection**_.
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Each element in the result must be**unique** and you may return the result in**any order**.
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Difficulty:**Easy**
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>Array Intersection: The intersection of two arrays is defined as the set of elements that are present in both arrays.
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###[Example 1]
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**Input**:`nums1 = [1,2,2,1], nums2 = [2,2]`
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-`0 <= nums1[i], nums2[i] <= 1000`
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##Intuition
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[中文题解](#中文题解)
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1. Convert one of the arrays to a`set`. The elements are unique in a`set`.
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2. When traversing the other array, if the an element is found to already exist in the`set`, it means that the element belongs to the intersection, and the element should be added to the`results`.
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3. The`results` is also of`set` type because duplicate removal is required.
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##Complexity
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* Time:`O(n)`.
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* Space:`O(n)`.
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* Time:`O(N)`.
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* Space:`O(N)`.
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##Java
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```java
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##问题描述
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给定两个数组`nums1` 和`nums2` ，返回_它们的**交集**_ 。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** 。
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难度:**容易**
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###[示例 1]
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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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###[示例 2]
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**输入**:`nums1 = [4,9,5], nums2 = [9,4,9,8,4]`
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**输出**:`[9,4]` 或者`[4,9]`
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#中文题解
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##思路
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1. 把其中一个数组转为`set`，数据结构`set`的特点是元素不重复。
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2. 遍历另一个数组时，如果发现当前元素已经存在于`set`中，则说明该元素属于交集，将该元素加入结果集中。
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3. 结果集也采用`set`类型，因为需要去重。

‎en/3001-4000/unorganized.md

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##other finished problems
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-https://leetcode.com/problems/k-closest-points-to-origin/
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-https://leetcode.com/problems/find-special-substring-of-length-k/
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-https://leetcode.cn/problems/eat-pizzas/
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‎zh/1-1000/349-intersection-of-two-arrays.md

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#349. Intersection of Two Arrays - Best Practices of LeetCode Solutions
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LeetCode link:[349. Intersection of Two Arrays](https://leetcode.com/problems/intersection-of-two-arrays),
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[349. 两个数组的交集](https://leetcode.cn/problems/intersection-of-two-arrays)
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#349. 两个数组的交集 - 力扣题解最佳实践
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力扣链接：[349. 两个数组的交集](https://leetcode.cn/problems/intersection-of-two-arrays) ，难度：**简单**。
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[中文题解](#中文题解)
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##力扣“349. 两个数组的交集”问题描述
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给定两个数组`nums1` 和`nums2` ，返回_它们的**交集**_ 。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**。
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##LeetCode problem description
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Given two integer arrays`nums1` and`nums2`, return_an array of their**intersection**_.
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Each element in the result must be**unique** and you may return the result in**any order**.
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>数组的交集: The intersection of two arrays is defined as the set of elements that are present in both arrays.
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Difficulty:**Easy**
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###[示例 1]
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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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###[Example 1]
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**Input**:`nums1 = [1,2,2,1], nums2 = [2,2]`
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###[示例 2]
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**Output**:`[2]`
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**输入**:`nums1 = [4,9,5], nums2 = [9,4,9,8,4]`
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###[Example 2]
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**Input**:`nums1 = [4,9,5], nums2 = [9,4,9,8,4]`
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**输出**:`[9,4]`
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**Output**:`[9,4]` or`[4,9]`
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**解释**:`[4,9] 也是可通过的`
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###[Constraints]
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###[约束]
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-`1 <= nums1.length, nums2.length <= 1000`
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-`0 <= nums1[i], nums2[i] <= 1000`
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##Intuition
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[中文题解](#中文题解)
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1. Convert one of the arrays to a`set`. The elements are unique in a`set`.
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2. When traversing the other array, if the an element is found to already exist in the`set`, it means that the element belongs to the intersection, and the element should be added to the`results`.
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3. The`results` is also of`set` type because duplicate removal is required.
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##思路
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1. 把其中一个数组转为`set`，数据结构`set`的特点是元素不重复。
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2. 遍历另一个数组时，如果发现当前元素已经存在于`set`中，则说明该元素属于交集，将该元素加入结果集中。
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3. 结果集也采用`set`类型，因为需要去重。
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##Complexity
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*Time:`O(n)`.
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*Space:`O(n)`.
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##复杂度
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*时间：`O(N)`。
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*空间：`O(N)`。
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##Java
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```java
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end
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```
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##C
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```c
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Kotlin
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```kotlin
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Swift
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```swift
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Rust
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```rust
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##Other languages
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##C, Kotlin, Swift, Rust or other languages
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```
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// Welcome to create a PR to complete the code of this language, thanks!
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```
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##问题描述
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给定两个数组`nums1` 和`nums2` ，返回_它们的**交集**_ 。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** 。
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难度:**容易**
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###[示例 1]
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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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###[示例 2]
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**输入**:`nums1 = [4,9,5], nums2 = [9,4,9,8,4]`
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**输出**:`[9,4]` 或者`[4,9]`
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#中文题解
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##思路
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1. 把其中一个数组转为`set`，数据结构`set`的特点是元素不重复。
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2. 遍历另一个数组时，如果发现当前元素已经存在于`set`中，则说明该元素属于交集，将该元素加入结果集中。
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3. 结果集也采用`set`类型，因为需要去重。

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