LeetCode_ 4 sum

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:
Elements in a quadruplet (a,b,c,d) must be in non-descending order. (ie, a ≤ b ≤ c ≤ d)
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)

　　无聊 3sum 的变形

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Note: The Solution object is instantiated only once and is reused by each test case.
        vector<vector<int>> res;
        int len = num.size();
        if(len < ) return res;
        sort(num.begin(),num.end());
        for(int i = ; i< len-;++i){
            while(i> && i< len- && num[i] == num[i-])++i;
            for(int j = i+; j< len-; ++j){
                while(j!=i+ && j< len-&&num[j] == num[j-])++j;
                int left = j+;
                int right = len-;
                while(left < right){
                    int sum = num[i] + num[j] +num[left]+num[right];
                    if(sum == target){
                        vector<int> ans;
                        ans.push_back(num[i]);
                        ans.push_back(num[j]);
                        ans.push_back(num[left]);
                        ans.push_back(num[right]);
                        res.push_back(ans);
                        left++;
                        while(left<right && num[left]==num[left-]) ++left;
                        right--;
                        while(left < right && num[right] == num[right+]) --right;
                    }else if(sum <target){
                         left++;
                        while(left<right && num[left]==num[left-]) ++left;
                    }else{
                         right--;
                        while(left < right && num[right] == num[right+]) --right;
                    }
                }//while(left <right)
            }//for j
        }// for i
        return res;
    }
};