题目：
Given an array of integers, find two numbers such that they add up to a specific target number.
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
代码：
class Solution { public: vector<int> twoSum(vector<int> &numbers, int target) { std::vector<int> ret_vector; std::map<int,int> value_index; for (int i = 0; i < numbers.size(); ++i) { const int gap = target - numbers[i]; if (value_index.find(gap) != value_index.end()) { ret_vector.push_back(std::min(i+1,value_index[gap]+1)); ret_vector.push_back(std::max(i+1,value_index[gap]+1)); break; } else { value_index[numbers[i]] = i; } } return ret_vector; } };
Tips:
元素无序且要求复杂度O(n)的，就可以用hashmap解决。
网上有的算法先遍历一遍numbers获得所有元素的map<value,index>，再进行后续的计算。这样的算法没有考虑数组元素重复的case
比如:
numbers = [0,2,4,0]
target = 0
原文：http://www.cnblogs.com/xbf9xbf/p/4437052.html