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].
思路：用一个map存储数组的下标后，已知一个数以及两数之和，在表中查找另一个数，因为假设输入只有一个答案，所以下标交换没有意义。
class Solution { public: vector<int> twoSum(vector<int>& nums, int target) { map<int, int> mapping; vector<int> result; for(int i=0;i<nums.size();i++) { mapping[nums[i]]=i; } for(int i=0;i<nums.size();i++) { int gap=target-nums[i]; if(mapping.find(gap)!=mapping.end()&&mapping[gap]>i) { result.push_back(i); result.push_back(mapping[gap]); break; } } return result; } };