# LeetCode 18. 4Sum

Given an array `nums` of *n* integers and an integer `target`, are there elements *a*, *b*, *c*, and *d* in `nums` such that *a* + *b* + *c* + *d* = `target`? Find all unique quadruplets in the array which gives the sum of `target`.

**Notice** that the solution set must not contain duplicate quadruplets.

**Example 1:**

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

**Example 2:**

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

**Constraints:**

* `0 <= nums.length <= 200`
* `-10^9 <= nums[i] <= 10^9`
* `-10^9 <= target <= 10^9`

## Solution:

[English Version in Youtube](https://youtu.be/zM8yfKADVfM)

[中文版解答Youtube Link](https://youtu.be/HJ8rgwmO1L0)

[中文版解答Bilibili Link](https://www.bilibili.com/video/BV1by4y1J7GF/)

```
class Solution {
    
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        vector<vector<int>>result;
        int n = num.size();
        sort(num.begin(), num.end());
        
        for (int i = 0; i < n; i++) {
            if (i > 0 && num[i-1] == num[i]) {
                continue;
            }
            for (int j = i + 1; j < n; j++) {
                if (j > (i + 1) && num[j-1] == num[j]) {
                    continue;
                }
                int l = j + 1, r = n - 1;
                int partial_sum = num[i] + num[j];
                while (l < r) {
                    int sum = partial_sum + num[l] + num[r];
                    if (sum < target) {
                        l++;
                    } else if (sum > target) {
                        r--;
                    } else {
                        result.push_back({num[i], num[j], num[l], num[r]});
                        while ((l + 1) < num.size() && num[l] == num[l+1]) l++;
                        l++;
                        while ((r - 1) >= 0 && num[r] == num[r - 1]) r--;
                        r--;
                    }
                }
            }
        }
        return result;
    }
};
```


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