两数之和与目标值对比，相同则返回这两个数的数组下标。

//时间复杂度为O(n^2)
#include<iostream>
#include<string>
using namespace std;
class Solution {
public:
    int* twoSum( int arr[],int nums, int target) {
        for (int k = 0; k < nums; k++)
        {
            for (int j = k + 1; j < nums; j++)
            {
                int temp = arr[k] + arr[j];
                if (temp == target)
                {
                    int xb[2] = { k,j };
                    cout<<"The subscripts of these two numbers are:" ;
                    cout<<xb[0]<<","<<xb[1];
                    return xb;
                }

            }
        }
    }
};
int main()
{
    Solution d;
    int arr[] = { 11,7,2,15 };
    //cout<<arr[0]<<arr[1];
    int nums=4;
    int target=9;

    int * p = d.twoSum(arr,nums,target);


    
}