Description
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
分析
题目的意思是：给定一个数组，求出最长的非负乘积的子数组。首先子数组不包含0，所以可以按照0进行讨论，我参考了一下别人的实现。
- pos_len记录乘积为正的长度，neg_len记录的是乘积为负的长度。
- 遍历数组，如果当前的数为0，则pos_len和neg_len都为0。如果当前的数是负数，则pos_len和neg_len需要交换了，因为负数使得前面的累积子数组正负性发生了变化，然后neg_len和pos_len的长度加1。如果当前的数为正，则neg_len和pos_len的长度加1。上述加1的过程要区分为0的情况，因为需要考虑到前面遍历的子数组包含0的情况哈。
- 遍历的时候维护更新res就能得到答案了。
代码
class Solution:
def getMaxLen(self, nums: List[int]) -> int:
pos_len=0
neg_len=0
res=0
for num in nums:
if(num==0):
pos_len,neg_len=0,0
elif(num<0):
pos_len,neg_len=neg_len,pos_len
neg_len+=1
if(pos_len>0):
pos_len+=1
else:
pos_len+=1
if(neg_len>0):
neg_len+=1
res=max(res,pos_len)
return res
参考文献
Leetcode 1567. Maximum Length of Subarray With Positive Product (python)
