从数组中任意取出2个数，判断他们的和是否为输入的数字sum，时间复杂度为0(n^2),空间复杂度0(1)

#include <stdio.h>
#include <stdlib.h>
#include <iostream>
using namespace std;

int a[] = {1,2,3,4,5,6,7,8,9,10};
int size = sizeof(a) / sizeof(int);
void twoSum(int data[], unsigned int length, int sum)
{
	int begin = 0;
	int end = length - 1;
	int currentSum = data[begin] + data[end];
	//假设已经排序好了
	while (begin < end)
	{
		if (currentSum == sum)
		{
			printf("%d\n,%d", data[begin] + data[end]);
			break;
		}
		else
		{
			if (currentSum < sum)
			{
				begin++;
			}
			else
			{
				end--;
			}
		}
	}
}