LeetCode in Python 172. Factorial Trailing Zeroes 阶乘后的零

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.

其实就是寻找n!表达式分解后5的个数，因为5*2=10，而2在5之前，个数一定比5多。而5的次方如25,125,625等含有多个5，连续除以5以计算这些多余的5。就是n/5+n/25+n/125...即n/5+n/5/5+n/5/5/5，如下的while循环：

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