Algorithms: Max Consecutive Ones

Challenge from LeetCode: count the maximum number of consecutive 1s in a list of 1s and 0s. I got it done pretty quickly, or so I thought. I had not accounted for what would happen upon the last iteration through the for loop. Reworked it and this is what I got: 

def checkMax(maxConsec, count):

        if count > maxConsec:

            return count, 0

        else:

            return maxConsec, 0

        

class Solution:    

    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:

        maxConsec = 0

        count = 0

       

        for index, num in enumerate(nums):

            if num == 1:  

                count += 1

                if index+1 == len(nums):

                    maxConsec, count = checkMax(maxConsec, count)

            else:

                maxConsec, count = checkMax(maxConsec, count)

                    

        return maxConsec

The memory usage of my code

 

Runtime. Oops.

I'm sure there are better ways to do this. I thought of putting all my counts in a list and checking for the largest value in the list once the for loop was complete. I wonder if that would make a difference to the runtime and memory usage?