Easy 1 Two Sum（1）

Description
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, and you may not use the same element twice.

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

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

Solution（c++）
给的nums是个无序的状态，所以用hash查找效率可以。(unordered_map的使用)

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        vector<int> res(2);        unordered_map<int,int> hash;        int length=nums.size();        for (int i=0;i<length;i++)           {              int NumToFind=target-nums[i];              if(hash.find(NumToFind)!=hash.end()){                  res[0]=hash[NumToFind];                  res[1]=i;                  return res;              }              hash[nums[i]]=i;           }    }};