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
[leetcode]Maximum Product Subarray @ Python
原文：http://www.cnblogs.com/zuoyuan/p/4019326.html