总结
- 善于变通
- 文章来源：LawsonAbs
- 待更新~
想法
想法一
01.改变正负号。如果是偶数个，则符号全部变正。
02.根据改变后的符号再来求最大积。
这种方法是有问题的：
[-1,-4,-3] 在遇到这种样例的时候，数组会变成 [1,4,-3] 这样，而不是[-1,4,3] ，然而后者才是正确的答案。
# class Solution:
# def maxProduct(self, nums: List[int]) -> int:
# pre_idx = None
# cnt = 0
# for i in range(len(nums)):
# if nums[i] < 0 : # 如果当前的值小于0
# if pre_idx is None:
# pre_idx = i
# cnt += 1
# elif nums[i] == 0: # 重置
# cnt = 0
# pre_idx = None
# if cnt % 2 == 0 and cnt: # 已经有连续两个负号
# nums[pre_idx] *= -1
# nums[i] *= -1 # 改符号
# pre_idx = None
# cnt = 0
# # product = copy(nums)
# print(nums)
# max_val = 0
# whole = 0
# for i in range(1,len(nums)):
# if nums[i] > 0 and nums[i-1] > 0:
# nums[i] = nums[i-1] * nums[i]
# whole += 1
# max_val = max(max_val,nums[i])
# if whole == len(nums): # 如果是全部数字都算进去了
# pass
# print(nums)
# print(max_val)
想法二
这里既然涉及到了正负号… 可AC代码如下
from typing import List
from copy import copy
class Solution():
def maxProduct(self, nums: List[int]) -> int:
dp_min = copy(nums) # 第i项的最小值
dp_max = copy(nums) # 第i项的最大值
total = nums[0]
for i in range(1,len(nums)):
total *= nums[i]
dp_min[i] = min(dp_min[i-1]*nums[i],dp_max[i-1]*nums[i],nums[i])
dp_max[i] = max(dp_min[i-1]*nums[i],dp_max[i-1]*nums[i],nums[i])
# print(dp_max)
res = max(dp_max)
return res
s = Solution()
nums = [-2,3,-2,4,-1,0,0,-4]
# nums = [-2,-1,-3,-4] # 得到的结果应该是12
# nums = [-1,-4,-3]
nums = [-2,-3,1] # 会计算整个数组的乘积
# nums = [-6,-2,-3,4,-5]
# nums = [-6,-2,-3,4]
# nums = [ -2,-3,4,-5]
out = s.maxProduct(nums)
print(out)
