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].

寻找两数之和，利用map存储数组的值与索引，然后遍历数组找出map中的元素，返回元素的索引。需要注意的是返回的结果索引不能相等。

class Solution {public:    vector
     
       twoSum(vector
      
       & nums, int target) {        unordered_map
       
         m;        for (int i = 0; i != nums.size(); i++)            m[nums[i]] = i;        for (int i = 0; i != nums.size(); i++) {            int diff = target - nums[i];            if (m.count(diff) && m[diff] != i)                return {i, m[diff]};        }    }};// 6 ms