#!/usr/bin/python
# -*- coding: utf-8 -*-
'''
Maximum Product Subarray.最大乘積的連續子數組
Find the contiguous subarray within an array (containing at least one number) which has the largest product.
For example, given the array [2,3,-2,4],
the contiguous subarray [2,3] has the largest product = 6.
'''
class Solution(object):
def maxProduct(self, nums):
"""
這種方法超過時間限制
:type nums: List[int]
:rtype: int
"""
length = len(nums)
if length == :
return
max = nums[]
for index,val in enumerate(nums):
sum =
for val2 in nums[index:]:
sum = sum * val2
if sum > max:
max = sum
return max
'''其實子數組乘積最大值的可能性為:累乘的最大值碰到了一個正數;或者,累乘的最小值(負數),碰到了一個負數。是以每次要儲存累乘的最大(正數)和最小值(負數)。
同時還有一個選擇起點的邏輯,如果之前的最大和最小值同目前元素相乘之後,沒有目前元素大(或小)那麼目前元素就可作為新的起點。
例如,前一個元素為0的情況,{1,0,9,2},到9的時候9應該作為一個最大值,也就是新的起點,{1,0,-9,-2}也是同樣道理,-9比目前最小值還小,是以更新為目前最小值。'''
def maxProduct(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
length = len(nums)
if length == :
return
max_tmp = min_tmp = ret = nums[]
for val in nums[:]:
#print max_tmp,min_tmp
a = max_tmp * val
b = min_tmp * val
max_tmp = max(a,b,val)
min_tmp = min(a,b,val)
ret = max(ret,max_tmp)
return ret
if __name__ == "__main__":
s = Solution()
print s.maxProduct([,,-,])
print s.maxProduct([,,-,-])
print s.maxProduct([,,,])
print s.maxProduct([,,,])
print s.maxProduct([,-,-,-,])
參考連結:http://blog.csdn.net/worldwindjp/article/details/39826823