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

Example 1:

Input: 3
Output: 0
Explanation: 3! = 6, no trailing zero.

Example 2:

Input: 5
Output: 1
Explanation: 5! = 120, one trailing zero.

Note: Your solution should be in logarithmic time complexity.

Python

 
class Solution(object):
    def trailingZeroes(self, n):
        """
        :type n: int
        :rtype: int
        """
        if n == 0:
            return 0
        else:
            return n/5 + self.trailingZeroes(n/5)

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