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].

Boring brushing leetcode.
Found that the subject is a company interview topic.

This problem mainly use unordered_map .

unordered_map allow two or more apper in data.but is  not ordered.

this is  my solutation:

class Solution {public:    vector<int> twoSum(vector<int>& nums, int target) {        unordered_map<int, int> mp;        vector<int> v;        for (int i = 0; i < nums.size(); ++i ){            int  x = target - nums[i];            if (mp.count(x) > 0) {                v.push_back((mp.find(x))->second);                v.push_back(i);                break;            }            mp.insert({nums[i],i});        }        return v;    }};

o(n)

1. Two Sum