One : subject

Two : Upper code

class Solution {
    
public:
    /**  Ideas : 1. The difference between this and the sum of three is that this question adds a layer of circulation , That is, we add the second layer loop after the first element we traverse  2. Other ideas are the same as the sum of three  */

    vector<vector<int>> fourSum(vector<int>& nums, int target) {
    
        
        vector<vector<int> >ans;
        sort(nums.begin(),nums.end());

        for (int i = 0; i < nums.size(); i++) {
    
            
            if (i > 0 && nums[i] == nums[i-1]) continue;// To reprocess 
                                                        // It can't be  nums[i] = nums[i+1]
                                                        // Because then we'll miss it  -1 -1 1 1 taget = 0
            for (int j = i+1; j < nums.size(); j++) {
    

                if ( j > i+1 && nums[j] == nums[j-1]) continue;

                int left = j+1;
                int right = nums.size()-1;

                while (left < right) {
    
                    
                    if (nums[i]+nums[j] > target - (nums[left]+nums[right])) {
    
                        right--;
                    } else if (nums[i]+nums[j] < target - (nums[left]+nums[right])) {
    
                        left++;
                    } else {
    
                        
                        ans.push_back({
    nums[i],nums[j],nums[left],nums[right]});

                        while (left < right && nums[left] == nums[left+1]) left++;
                        while (left < right && nums[right] == nums[right-1]) right--;

                        // The top de duplication is only the de duplication to the previous element of different elements .
                        left++;
                        right--;
                    }
                }
            }
        }

        return ans;

    }
};