Question：

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)

Anwser 1：

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        
        vector< vector<int> > ret;
        
        int len = num.size();
        if(len < 3) return ret;
        
        vector<int> num2;
        num2.assign(num.begin(), num.end());
        
        sort( num2.begin(), num2.end() );       // sort  O(n*log(n))
        
        
        vector<int> tmpRet(3);
        tmpRet[0] = num2[0] - 1;
        
        for(int i = 0; i < len - 2; i++){       // O(n*m)
            int l = i + 1;
            int r = len - 1;
            
            if(num2[i] == tmpRet[0]) continue;
            tmpRet[0] = num2[i];
            
            while(l < r) {
                int sum2 = num2[l] + num2[r];
                if(num2[i] + sum2 == 0){
                    tmpRet[1] = num2[l];
                    tmpRet[2] = num2[r];
                    
                    ret.push_back(tmpRet);
                    
                    while(num2[++l] == num2[l-1] && l < r);   // remove same number
                    while(num2[--r] == num2[r+1] && l < r);  
            
                } else if(num2[i] + sum2 < 0){
                    l++;
                } else {
                    r--;
                }
            }
        }
        
        return ret;
    }
};

原文： 【leetcode】3Sum