Two Sum ——经典的哈希表的题

Given an array of integers, return indices of the two numbers such that they add up to a specific target.

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

Example:

Given nums = [2, 7, 11, 15], target = 9,

Because nums[0] + nums[1] = 2 + 7 = 9,
return [0, 1].

UPDATE (2016/2/13):
The return format had been changed to zero-based indices. Please read the above updated description carefully.

思路是循环一次，每次都判断当前数组索引位置的值在不在hashtable里，不在的话，加入进去，key为数值，value为它的索引值；在的话，取得他的key，记为n（此时n一定小于循环变量i），接下来再在hashtable中查找（target-当前数值）这个数，利用了hashtable中查找元素时间为常数的优势，如果找到了就结束，此处需要注意的是，如果数组中有重复的值出现，那么第二次出现时就不会加入到hashtable里了，比如3,4,3,6；target=6时，当循环到第二个3时，也可以得到正确结果。

        最终ac的代码如下：

class Solution {
public:
    vector<int> twoSum(vector<int> &numbers, int target) {
        vector<int > res;
        map<int,int> numMap;
        map<int,int>::iterator iter;
        int len=numbers.size();
        for(int i=;i<len;i++)
        {
            iter=numMap.find(target-numbers[i]);
            if(iter!=numMap.end())
            {
                res.push_back(iter->second);
                res.push_back(i);
                return res;
            }
            else
            {
                numMap[numbers[i]]=i;
            }
        }

    }
};