输入一个递增排序的数组和一个数字S，在数组中查找两个数，使得他们的和正好是S，如果有多对数字的和等于S，输出两个数的乘积最小的。

// test20.cpp : 定义控制台应用程序的入口点。

//

#include "stdafx.h"
#include<iostream>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<cstring>
#include<string.h>
#include<deque>

using namespace std;

class Solution {
public:
	vector<int> FindNumbersWithSum(vector<int> array, int sum) {
		if (array.empty()) return{};
		auto small = array.begin(); //指向小的指针
		auto big = array.end()-1; //指向大的指针
		auto curr = big;
		vector<vector<int>> vec;
		vector<int> result;

		while (small<big)
		{
			while (small < curr)
			{
				if (*small + *curr == sum)  //找到第一组就能确定是乘积最小，所以不用做后续的比较
				{
					result.push_back(*small);
					result.push_back(*curr);
					vec.push_back(result);
					result.clear();
				}
				else if (*small + *curr < sum)
				{
					break;
				}
				else
				{

				}
				--curr;
			}
			small++;
			curr = big;
		}
		if (vec.empty()) return{};
		auto it = vec.begin();//这里面是多余的
		int num1= *(it->begin());
		int num2= *(it->begin() + 1);
		int mult = num1*num2;
		++it;
		while (it!=vec.end())
		{
			if (mult >= (*(it->begin()))*(*(it->begin() + 1)))
			{
				num1 = *(it->begin());
				num2 = *(it->begin() + 1);
			}
			++it;
		}
		result.clear();
		result.push_back(num1);
		result.push_back(num2);

		return result;
	}
};
int main()
{

	Solution so;
	vector<int> v = { 1,2,3,4,5,6,7,8,9 };
	vector<int> vec = so.FindNumbersWithSum(v,9);
	for (auto it = vec.begin();it != vec.end();++it)
	{
		cout << *it << "  ";
	}

	cout << endl;
	return 0;
}