题目描述： 
The function twoSum should return indices of the two numbers such that they add up to the target, where index1 must be less than index2. Please note that your returned answers (both index1 and index2) are not zero-based.

You may assume that each input would have exactly one solution.

Input: numbers={2, 7, 11, 15}, target=9 
Output: index1=1, index2=2 
可能出现的badcase: 答案中存在一个数为target/2时。 
实现思路：若第一层遍历数1，第二层遍历数2，则时间复杂度O(n2)遍历。若仅进行第一遍遍历，在寻找第二个数的时候使用Hash表存储数据，则时间复杂度为O(n)，此时需要注意badcase问题。

实现代码如下：

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        map<int, int> hash_map;
        vector<int> r;
        for (int i=0; i != numbers.size(); ++i)
        {
            hash_map[numbers[i]] = i;
        }
        for(int i=0; i != numbers.size(); ++i)
        {
            map<int, int>::iterator it = hash_map.find(target-numbers[i]);
            if (it != hash_map.end())
            {
                if(i+1 == it->second+1)
                    continue;
                r.push_back(min(i+1, it->second+1));
                r.push_back(max(i+1, it->second+1));
            }
        }
        return r;
    }
};

LeetCode实践之一 Two Sum