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)