18. 4Sum -- 找到数组中和为target的4个数

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

Note: The solution set must not contain duplicate quadruplets.

For example, given array S = [1, 0, -1, 0, -2, 2], and target = 0.

A solution set is:
[
  [-1,  0, 0, 1],
  [-2, -1, 1, 2],
  [-2,  0, 0, 2]
]

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        int l = nums.size(), sum, left, right, i, j;
        vector<vector<int>> ans;
        if (l<4)
            return ans;
        sort(nums.begin(), nums.end());
        for (i = 0; i<l - 3; i++)
        {
            if(i && nums[i] == nums[i-1])
                continue;
            for (j = i + 1; j<l - 2; j++)
            {
                if(nums[i]+nums[j]+nums[j+1]+nums[j+2]>target)
                    break;
                if(nums[i]+nums[j]+nums[l-2]+nums[l-1]<target)
                    continue;
                if (j>i + 1 && nums[j] == nums[j - 1])
                    continue;
                sum = nums[i] + nums[j];
                left = j + 1;
                right = l - 1;
                while (left < right)
                {
                    if (sum + nums[left] + nums[right] == target)
                        ans.push_back({ nums[i], nums[j], nums[left++], nums[right--] });
                    else if (sum + nums[left] + nums[right] < target)
                        left++;
                    else
                        right--;
                    while (left>j+1 && nums[left] == nums[left - 1])
                        left++;
                    while(right<l-1 && nums[right] == nums[right + 1])
                        right--;
                }
            }
        }
        return ans;
    }
};