twosum

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].

//这不是最优方法，用哈希是最优的，见leetcode

#include<iostream>
#include<vector>
using namespace std;
class solution{
public:
vector<int> find(vector<int>&num,int&target)//由于这里的rev是在类的函数内声明的
//当函数返回时rev在返回后会销毁，因此不能返回引用，只能是值
{
vector<int> rev;
for (int i = 0; i < num.size(); i++)
{
for (int j = i+1; j < num.size(); j++)
{
int re = num[i] + num[j];
if (re == target)
{
rev.push_back(i);
rev.push_back(j);
return rev;
}
}
}

}
};
int main()
{
vector<int> num = { 2, 7, 11, 15 ,32,34,43,343,65};
int target = 77;
solution s;
vector<int> result=s.find(num, target);
cout << result[0] << ' ' << result[1] << endl;
}