【leetcode】1567. Maximum Length of Subarray With Positive Product

题目如下：

Given an array of integers nums, find the maximum length of a subarray where the product of all its elements is positive.

A subarray of an array is a consecutive sequence of zero or more values taken out of that array.

Return the maximum length of a subarray with positive product.

Example 1:
Input: nums = [1,-2,-3,4]
Output: 4
Explanation: The array nums already has a positive product of 24.
Example 2:
Input: nums = [0,1,-2,-3,-4]
Output: 3
Explanation: The longest subarray with positive product is [1,-2,-3] which has a product of 6.
Notice that we cannot include 0 in the subarray since that'll make the product 0 which is not positive.
Example 3:
Input: nums = [-1,-2,-3,0,1]
Output: 2
Explanation: The longest subarray with positive product is [-1,-2] or [-2,-3].
Example 4:
Input: nums = [-1,2]
Output: 1
Example 5:
Input: nums = [1,2,3,5,-6,4,0,10]
Output: 4
Constraints:

1 <= nums.length <= 10^5

-10^9 <= nums[i] <= 10^9

解题思路：根据题目要求，0是不能被计算在子数组之内的，因此可以理解成0比nums分成了若干个子数组。对于每个子数组，我们只需要从左往右遍历，记录当前负数的个数以及第一次出现负数为奇数时候的下标。这样的话，如果当前负数的个数是偶数，那么这一段子数组的乘积为正数；而如果是奇数，则与第一次出现负数为奇数时组成的这一段子数组中有偶数个负数，其成绩也为正数。

代码如下：

class Solution(object):
    def getMaxLen(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        last_neg_odd = None
        neg_count = 0

        start = None

        res = 0

        for i in range(len(nums)):
            if nums[i] == 0:
                last_neg_odd = None
                neg_count = 0
                start = None
                continue
            elif start == None:
                start = i

            if nums[i] < 0:neg_count += 1

            if neg_count % 2 == 0:
                res = max(res,i - start + 1)
            else:
                if last_neg_odd == None:
                    last_neg_odd = i
                else:
                    res = max(res, i - last_neg_odd)


        return res