【优选算法】四数之和（双指针算法）

class Solution {
public:
    vector<vector<int>> fourSum(vector<int>& nums, int target) {
        vector<vector<int>> ret;
        int n = nums.size();
        //1.排序
        sort(nums.begin(), nums.end());

        //2.先固定一个数a，在剩下的数里面利用三数之和等于target-a
        for (int i = 0; i < n; )
        {
            //先固定一个数b，在剩下的数里面利用双指针算法
            for (int j = i + 1; j < n; )
            {
                int left = j + 1, right = n - 1;
                //避免数据溢出，我们用long long类型
                long long count = (long long)target - nums[i] - nums[j];
                while (left < right)
                {
                    int sum = nums[left] + nums[right];
                    if (sum > count) right--;
                    else if (sum < count) left++;
                    else
                    {
                        ret.push_back({ nums[i],nums[j],nums[left++],nums[right--] });
                        //去重一，left和right
                        while (left < right && nums[left] == nums[left - 1]) left++;
                        while (left < right && nums[right] == nums[right + 1]) right--;
                    }
                }
                //去重二
                j++;
                while (j < n && nums[j] == nums[j - 1]) j++;
            }
            //去重三
            i++;
            while (i < n && nums[i] == nums[i - 1]) i++;
        }
        return ret;
    }