学步园

题目来源：http://www.leetcode.com/onlinejudge

解题报告：

简单的用递归实现的，有一点是注意数组中可能会有重复数字，所以要考虑结果中不要输出重复结果。

#include <iostream>
#include <algorithm>
#include <vector>


using namespace std;

class Solution {
public:
    
    void f(int index, vector<int> temp, vector<int> &num, vector<vector<int> > &result, int sum)
    {
        if (temp.size() == 3)
        {
            if (sum==0)
                result.push_back(temp);
            return;
            
        }
        if (index == num.size())
            return;
        if (sum + num[index] > 0)
            return;
        int i = index;

        int next = i+1;
        while(next < num.size() && num[next] == num[i])
            next++;
        temp.push_back(num[index]);
        f(i+1, temp, num, result, sum+num[index]);
        temp.pop_back();
        f(next, temp, num, result, sum);

    }
    
    vector<vector<int> > threeSum(vector<int> &num) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        sort(num.begin(),num.end());
        vector<vector<int> > result;
        vector<int> temp;
        f(0,temp,num,result,0);
        return result;
    }
    

};

附录：

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)