原题地址：https://oj.leetcode.com/problems/maximum-product-subarray/
解题思路：主要需要考虑负负得正这种情况，比如之前的最小值是一个负数，再乘以一个负数就有可能成为一个很大的正数。
代码：
class Solution:
# @param A, a list of integers
# @return an integer
def maxProduct(self, A):
if len(A) == 0:
return 0
min_tmp = A[0]
max_tmp = A[0]
result = A[0]
for i in range(1, len(A)):
a = A[i] * min_tmp
b = A[i] * max_tmp
c = A[i]
max_tmp = max(max(a,b),c)
min_tmp = min(min(a,b),c)
result = max_tmp if max_tmp > result else result
return result