输出递增数组中和为sum的两个数

算法本身不复杂，但是提供了一个跟有序数组相关的一个思路：记录首尾指针，由两边向中间缩进。

#include <iostream> 
using namespace std;

bool findPairWithSum(int* a, int len, int sum, int& n1, int& n2)
{
	if(a==NULL || len<2 || sum<a[0]+a[1] || sum>a[len-1]+a[len-2])
		return false;

	int low=0, high=len-1;
	while(low<high)
	{
		if(a[low]+a[high] == sum)
		{
			n1 = a[low];
			n2 = a[high];
			return true;
		}else if(a[low]+a[high] > sum)
			high--;
		else
			low++;
	}
	return false;
}

int main()
{ 
	int a[] = {1,3,6,8,12,23,45,59,70,98,100};
	int len = sizeof(a)/sizeof(a[0]);
	int n1, n2;
	bool found;
	int sum = 68;
	found = findPairWithSum(a, len, sum, n1, n2);
	if(found)
		cout<<n1<<" "<<n2<<endl;
	return 0;
}