---
description: Two Pointer
---

# LeetCode 15. 3Sum



Given an array `nums` of _n_ integers, are there elements _a_, _b_, _c_ in `nums` such that _a_ + _b_ + _c_ = 0? Find all unique triplets in the array which gives the sum of zero.

Notice that the solution set must not contain duplicate triplets.

**Example 1:**

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

**Example 2:**

```
Input: nums = []
Output: []
```

**Example 3:**

```
Input: nums = [0]
Output: []
```

**Constraints:**

* `0 <= nums.length <= 3000`
* `-10^5 <= nums[i] <= 10^5`

## Solution:

[English Version in Youtube](https://youtu.be/3Szdf-_uRRE)

[中文版解答Youtube Link](https://youtu.be/v86W0cgvO5Y)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1cy4y1J71A/)

```
class Solution {
public:
    vector<vector<int>> threeSum(vector<int>& nums) {
        vector<vector<int>> result;
            
        sort(nums.begin(), nums.end());
        
        for (int i = 0; i < nums.size(); i++) {
            if (i > 0 && nums[i - 1] == nums[i]) {
                continue;
            }
            
            int j = i + 1;
            int k = nums.size() - 1;
            
            while (j < k) {
                int sum = nums[i] + nums[j] + nums[k];
                if (sum < 0) {
                    j++;
                } else if (sum > 0) {
                    k--;
                } else {
                    result.push_back({nums[i], nums[j], nums[k]});
                    while ((j + 1) < nums.size() && nums[j] == nums[j+1]) j++;
                    j++;
                    while ((k - 1) >= 0 && nums[k] == nums[k - 1]) k--;
                    k--;
                }
            }
        }
        
        return result;
    }
};
```
