3Sum

3Sum

Given an array S of n integers, are there elements a, b, c in S such that a + b + c = 0? 

Find all unique triplets in the array which gives the sum of zero.
Note:
Elements in a triplet (a,b,c) must be in non-descending order. (ie, a <= b <= c)
The solution set must not contain duplicate triplets.
For example, given array S = {-1 0 1 2 -1 -4},
A solution set is:
(-1, 0, 1)
(-1, -1, 2)

class Solution {
public:
    // sort and then search from both ends to middle
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > res;
        int N = num.size();
        if(N < 3) return res;
        sort(num.begin(), num.end());
       
        for(int k = N-1; k >= 2; k--) {
            if(k != N-1 && num[k] == num[k+1]) {
                continue; // handle duplicated cases 
            } 
            for(int i = 0, j = k-1; i < j; ) {
                if(i != 0 && num[i] == num[i-1]) { 
                    i++; 
                    continue; 
                }
                if(j != k-1 && num[j] == num[j+1]) { 
                    j--; 
                    continue; 
                }
                int sum = num[i] + num[j] + num[k];
                if(sum == 0) {
                    vector<int> tmp(3);
                    tmp[0] = num[i];
                    tmp[1] = num[j];
                    tmp[2] = num[k];
                    res.push_back(tmp);
                    i++; 
                    j--;
                }
                else if(sum < 0) { 
                    i++; 
                }
                else { 
                    j--; 
                }
            }
        }
        return res;
    }
};