子数组的最大乘积（编程之美2.13，两种解法）

#include "stdafx.h"

//书解法一
int max(int * arr, size_t len)
{
	int * s = new int[len+1];
	int * t = new int[len+1];
	int * p = new int[len];
	s[0] = 1;
	t[len] = 1;

	for(int i=1;i<=len;i++)
	{
		s[i] = s[i-1] * arr[i-1];
		t[len-i] = t[len-i+1] * arr[len-i];
	}
	int max = s[0] * t[1];

	for(int i=1;i<len;i++)
	{
		p[i] = s[i] * t[i+1];
		if(p[i] > max)
			max = p[i];
	}
	return max;
}

//书解法二
int max2(int * arr, size_t len)
{
	int t = 1;
	int nzeros = 0;
	int max_neg = 0;
	int min_pos = 0;
	for(int i=0;i<len;i++)
	{
		if(arr[i]==0)
			nzeros++;
		else
		{
			t*=arr[i];
			if(arr[i]<0)
			{
				if(max_neg==0 || arr[i]>max_neg)
					max_neg = arr[i];
			}
			else
			{
				if(min_pos==0 || arr[i]<min_pos)
					min_pos = arr[i];
			}
		}
	}
		
	if(nzeros > 1)
		return 0;
	else if(nzeros == 1)
	{
		if(t>0)
			return t;
		else
			return 0;
	}
	else if(t>0)
	{
		return t/min_pos;
	}
	else if(t<0)
	{
		return t/max_neg;
	}
	


}

int _tmain(int argc, _TCHAR* argv[])
{
	int arr[5] = {-2,-3,4,-1,6};
	printf("%d\n",max(arr,5));
	printf("%d\n",max2(arr,5));
	return 0;
}