[Leetcode 48] 18 4 Sum

Problem:

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)

Analysis:

It's very much like 3 Sum except that we need another variable start from the end of the given array

Code:

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
        vector<vector<int> > res;
        
        sort(num.begin(), num.end());
        
        int a = 0;
        while (a < num.size()) {
            int b = num.size()-1;
            while (b > a) {
                int c = a+1, d = b-1;
                
                while (c < d) {
                    int sum = num[a] + num[b] + num[c] + num[d];
                
                    if (sum == target) {
                        vector<int> tmp;
                        tmp.push_back(num[a]);
                        tmp.push_back(num[c]);
                        tmp.push_back(num[d]);
                        tmp.push_back(num[b]);
                    
                        res.push_back(tmp);
                    
                        do {c++;} while (num[c]==num[c-1] && c < b);
                        do {d--;} while (num[d]==num[d+1] && d > a);
                    } else if ( sum < target) {
                        c++;
                    } else {
                        d--;
                    }
                }
                
                do {b--;} while (num[b] == num[b+1] && b > 0);
            }
            
            do {a++;} while (num[a] == num[a-1] && a < num.size());
        }
        
        return res;
    }
};

Attention: