18. 4Sum

Given an array nums of n integers, return an array of all the unique quadruplets [nums[a], nums[b], nums[c], nums[d]] such that:

• 0 <= a, b, c, d < n
• a, b, c, and d are distinct.
• nums[a] + nums[b] + nums[c] + nums[d] == target

You may return the answer in any order.

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 = [2,2,2,2,2], target = 8
Output: [[2,2,2,2]]

Constraints:

• 1 <= nums.length <= 200
• -109 <= nums[i] <= 109
• -109 <= target <= 109

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

注意：

        1.数组越界问题：eg，需要先判断k>0，再判断nums[k]==nums[k-1]，否则会导致越界

        2.需要进行二重剪枝，但是这里和三数之和有不同的点，不是判断出第一个/第二个比target大就break，而是需要看是否为负数。