Leetcode--3Sum

Problem Description:

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)

分析：题目要求把全部可能的集合都找出来。因此想到的就是首先将数组排序，然后利用两重循环依次选出两个数字a和b。然后在剩下的数字中查找是否存在c，详细实现用到了upper_bound函数找到比当前数大的第一个数，去掉反复循环的情况。然后用find函数查找c是否存在，存在则将三个数记录下来。

详细代码例如以下：

class Solution {
public:
    vector<vector<int> > threeSum(vector<int> &num) {
        vector<vector<int> > results;
        if(num.size()<3) return results;
        vector<int> subset;
        sort(num.begin(),num.end());
        vector<int>::iterator p=num.begin(),q,flag;
        while(p<(num.end()-2))
        {
            q=p+1;
            while(q<num.end()-1)
            {
                int tag=0-*p-*q;
                if(find(q+1,num.end(),tag)!=num.end())
                {
                flag=find(q+1,num.end(),tag);
                subset.push_back(*p);
                subset.push_back(*q);
                subset.push_back(*flag);
                results.push_back(subset);
                }
                subset.clear();
                q=upper_bound(q,num.end()-1,*q);
            }
            p=upper_bound(p,num.end()-2,*p);
        }
        return results;
    }
};