1. Two Sum

1. Two Sum

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].

Analysis:
此题若使用O(n2)复杂度算法会超时，下面代码复杂度为O(nlogn)。
注意测试单元里要会两个数相同的情况。
参考：http://blog.csdn.net/jiadebin890724/article/details/23305449
Source Code（C++）:

#include <iostream>#include <vector>#include <map>using namespace std;class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        map<int, int> m;        vector<int> res;        map<int, int>::iterator it;        for(int i=0; i<nums.size(); i++) {            it = m.find(target-nums.at(i));            if (it != m.end()){                res.push_back(it->second);                res.push_back(i);                return res;            }            m[nums.at(i)]=i;        }        return res;//不加此句在leetcode上会出编译错误    }};int main() {    Solution sol;    vector<int> res;    vector<int> nums;    nums.push_back(2);    nums.push_back(7);    nums.push_back(2);    nums.push_back(15);    res = sol.twoSum(nums, 4);    for (int i=0; i<res.size(); i++){        cout << res.at(i);    }    return 0;}