【Easy】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].

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

中心思想： 

用unordered_map来存储数字和其出现的位置，每次存好后，查找一下数组中是不是存在(target-该数字)，如果存在，即可返回这两个数字的位置

注意易错的地方： 存hash和查找差值最好分开来做，因为如果遇到相同数字，其在hash map中的value会改变，变成后面那个的位置，此时已遍历经过了前一个数，就无法找出了

class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { unordered_map<int, int> hash; vector<int> result; if (nums.size() == 0) return result; for (int i = 0; i < nums.size(); i++){ hash[nums[i]] = i; } for (int i = 0; i < nums.size(); i++){ if (hash.count(target-nums[i]) && i!=hash[target-nums[i]] ){ result.push_back(i); result.push_back(hash[target-nums[i]]); return result; } } return result; } };