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45 lines
1.2 KiB
Plaintext
45 lines
1.2 KiB
Plaintext
---
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title: '15. 3Sum'
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description: 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
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sidebar:
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badge: 'Medium'
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---
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<Badge variant="accent">Two Pointers</Badge>
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::::warning
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Notice that the solution set must not contain duplicate triplets.
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::::
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### Example 1:
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- Input: `nums = [-1,0,1,2,-1,-4]`
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- Output: `[[-1,-1,2],[-1,0,1]]`
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- Explanation: `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.
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### Example 2:
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- Input: `nums = [0,1,1]`
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- Output: `[]`
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- Explanation: The only possible triplet does not sum up to `0`.
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### Example 3:
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- Input: `nums = [0,0,0]`
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- Output: `[[0,0,0]]`
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- Explanation: The only possible triplet sums up to `0`.
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### Constraints:
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- `3 <= nums.length <= 3000`
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- `-10^5 <= nums[i] <= 10^5`
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## Solution
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```js
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/**
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* @param {number[]} nums
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* @return {number[][]}
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*/
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var threeSum = function(nums) {
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};
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```
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