Factorial Trailing Zeroes

Given an integer n, return the number of trailing zeroes in n!.

Note: Your solution should be in logarithmic time complexity.

Concept:

Since 0 only company with 5*2 So only need to count the volume of 5 factor. (because 2 always enough)

So.. 100! 's zero has => floor(100/5) + floor(100/25) = floor(100/5) + floor((100/5)/5)

class Solution(object):
    def trailingZeroes(self, n):
        """
        :type n: int
        :rtype: int
        """
        res = 0
        while n > 0:
            n /= 5
            res += 1
        return res