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Description
tab: English
<p>Given an integer array nums, return all the triplets <code>[nums[i], nums[j], nums[k]]</code> such that <code>i != j</code>, <code>i != k</code>, and <code>j != k</code>, and <code>nums[i] + nums[j] + nums[k] == 0</code>.</p>
<p>Notice that the solution set must not contain duplicate triplets.</p>
<p> </p>
<p><strong class="example">Example 1:</strong></p>
<pre>
<strong>Input:</strong> nums = [-1,0,1,2,-1,-4]
<strong>Output:</strong> [[-1,-1,2],[-1,0,1]]
<strong>Explanation:</strong>
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0.
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0.
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0.
The distinct triplets are [-1,0,1] and [-1,-1,2].
Notice that the order of the output and the order of the triplets does not matter.
</pre>
<p><strong class="example">Example 2:</strong></p>
<pre>
<strong>Input:</strong> nums = [0,1,1]
<strong>Output:</strong> []
<strong>Explanation:</strong> The only possible triplet does not sum up to 0.
</pre>
<p><strong class="example">Example 3:</strong></p>
<pre>
<strong>Input:</strong> nums = [0,0,0]
<strong>Output:</strong> [[0,0,0]]
<strong>Explanation:</strong> The only possible triplet sums up to 0.
</pre>
<p> </p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>3 <= nums.length <= 3000</code></li>
<li><code>-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup></code></li>
</ul>
> [!tip]- Hint 1
>
> 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!
> [!tip]- Hint 2
>
> For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next number y, which is value - x where value is the input parameter. Can we change our array somehow so that this search becomes faster?
> [!tip]- Hint 3
>
> The second train of thought for two-sum is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?
---
[submissions](https://leetcode.com/problems/3sum/submissions/) | [solutions](https://leetcode.com/problems/3sum/solutions/)
tab: 中文
<p>给你一个整数数组 <code>nums</code> ，判断是否存在三元组 <code>[nums[i], nums[j], nums[k]]</code> 满足 <code>i != j</code>、<code>i != k</code> 且 <code>j != k</code> ，同时还满足 <code>nums[i] + nums[j] + nums[k] == 0</code> 。请你返回所有和为 <code>0</code> 且不重复的三元组。</p>
<p><strong>注意：</strong>答案中不可以包含重复的三元组。</p>
<p> </p>
<p> </p>
<p><strong>示例 1：</strong></p>
<pre>
<strong>输入：</strong>nums = [-1,0,1,2,-1,-4]
<strong>输出：</strong>[[-1,-1,2],[-1,0,1]]
<strong>解释：</strong>
nums[0] + nums[1] + nums[2] = (-1) + 0 + 1 = 0 。
nums[1] + nums[2] + nums[4] = 0 + 1 + (-1) = 0 。
nums[0] + nums[3] + nums[4] = (-1) + 2 + (-1) = 0 。
不同的三元组是 [-1,0,1] 和 [-1,-1,2] 。
注意，输出的顺序和三元组的顺序并不重要。
</pre>
<p><strong>示例 2：</strong></p>
<pre>
<strong>输入：</strong>nums = [0,1,1]
<strong>输出：</strong>[]
<strong>解释：</strong>唯一可能的三元组和不为 0 。
</pre>
<p><strong>示例 3：</strong></p>
<pre>
<strong>输入：</strong>nums = [0,0,0]
<strong>输出：</strong>[[0,0,0]]
<strong>解释：</strong>唯一可能的三元组和为 0 。
</pre>
<p> </p>
<p><strong>提示：</strong></p>
<ul>
<li><code>3 <= nums.length <= 3000</code></li>
<li><code>-10<sup>5</sup> <= nums[i] <= 10<sup>5</sup></code></li>
</ul>
> [!tip]- 提示 1
>
> 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!
> [!tip]- 提示 2
>
> For the two-sum problem, if we fix one of the numbers, say x, we have to scan the entire array to find the next number y, which is value - x where value is the input parameter. Can we change our array somehow so that this search becomes faster?
> [!tip]- 提示 3
>
> The second train of thought for two-sum is, without changing the array, can we use additional space somehow? Like maybe a hash map to speed up the search?
---
[提交记录](https://leetcode.cn/problems/3sum/submissions/) | [题解](https://leetcode.cn/problems/3sum/solution/)
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