LeetCode problem link:[15. 3Sum](https://leetcode.com/problems/3sum),

[15. 三数之和](https://leetcode.cn/problems/3sum)

LeetCode link:[15. 3Sum](https://leetcode.com/problems/3sum), difficulty:**Medium**

[中文题解](#中文题解)

Given an integer array nums, return all the triplets`[nums[i], nums[j], nums[k]]` such that`i != j`,`i != k`, and`j != k`, and`nums[i] + nums[j] + nums[k] == 0`.

Notice that the solution set must not contain duplicate triplets.

Difficulty:**Medium**

**Input**:`nums = [-1,0,1,2,-1,-4]`

<details>

<summary>Hint 1</summary>

So, we essentially need to find three numbers x, y, and z such that they add up to the given value. If we fix one of the numbers say x, we are left with the two-sum problem at hand!

</details>

[中文题解](#中文题解)

1. The`sum` of three numbers equals`0`, which is equivalent to the`sum` of_two numbers_ equaling the_**negative** third number_.

2. There are two options:

1. First`determine two numbers` and`find the third number`

* Time:`O(N * N)`.

* Space:`O(N)`.

[15. 三数之和](https://leetcode.cn/problems/3sum), 难度:**中等**。

给你一个整数数组`nums` ，判断是否存在三元组`[nums[i], nums[j], nums[k]]` 满足`i != j`、`i != k` 且`j != k` ，同时还满足`nums[i] + nums[j] + nums[k] == 0` 。请你返回所有和为`0` 且不重复的三元组。

**注意**：答案中**不**可以包含**重复**的三元组。

**输入**:`nums = [-1,0,1,2,-1,-4]`

**输出**:`[[-1,-1,2],[-1,0,1]]`

**解释**:

1. 三个数相加等于0，等同于`两个数相加的和`等于`负的第三个数`。

2. 有两种方案可供选择：

1. 先定下两个数，查找第三个数

2. 先定下一个数，查找另外两个数

3. 如果选择`方案1`，需要用到`Map`。因为需要对`nums`去重复；在`Map`中查找`第三个数`时，还需要避开已经定下的`两个数`，实现起来会比较麻烦。

4. 如果选择`方案2`，查找另外两个数时，需要用到`two pointers`算法。

5. 对于`方案1`，仅给出了`Python`示例代码。下文重点讲`方案2`。

1. 对`nums`进行排序。

2. 遍历`nums`。

3. 伪代码：

for (i=0; i<nums.length; i++) {

left= i+1

right=nums.length-1

while (left< right) {

if (condition1) {

left+=1

}else (condition2) {

right-=1

}

}

}

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