Question：

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)

Anwser 1：

O（n^3)

class Solution {
public:
    vector<vector<int> > fourSum(vector<int> &num, int target) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int len = num.size();
        vector< vector<int> > ret;
        
        if(len < 4) return ret;
        
        vector<int> num2;
        num2.assign(num.begin(), num.end());
        
        sort(num2.begin(), num2.end());
        
        vector<int> tmpRet(4);
        
        tmpRet[0] = num2[0] - 1;
        for(int i = 0; i < len - 3; i++){
            if(tmpRet[0] == num2[i]) continue;
            
            tmpRet[0] = num2[i];
            
            tmpRet[1] = num2[i+1] - 1;
            for(int j = i + 1; j < len - 2; j++){
                if(tmpRet[1] == num2[j]) continue;
                tmpRet[1] = num2[j];
                
                int l = j + 1;
                int r = len - 1;
                
                while(l < r){
                    if(tmpRet[0] + tmpRet[1] + num2[l] + num2[r] == target){
                        tmpRet[2] = num2[l];
                        tmpRet[3] = num2[r];
                        
                        ret.push_back(tmpRet);
                        
                        while(num2[++l] == num2[l-1] && l < r);
                        while(num2[--r] == num2[r+1] && l < r);
                        
                    } else if(tmpRet[0] + tmpRet[1] + num2[l] + num2[r] < target){
                        l++;
                    } else {
                        r--;
                    }
                }
            }
        }
        
    }
};

参考推荐：

3Sum

LeetCode: 4Sum