172. Factorial Trailing Zeroes

题目

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

Note: Your solution should be in logarithmic time complexity.

思路

• 后面的0是2*5产生的， 因为2的因子数量永远比5多，所以我们只要找有多少个5就好了
• 像25这种数，里面有两个5的因子，怎么办？ 举个例子，100！里面，有一个5的因子数的个数是100/5 = 20, 有两个5的因子个数是100/25 = 4。所以一共有24个。所以我们就可以用100一直除以5，把所得数相加。

Python

递归

class Solution(object):
    def trailingZeroes(self, n):
        """
        :type n: int
        :rtype: int
        """
        #A very smart question, 2 is always ample, so we only need to care about the factor of 5
        return  0 if n == 0 else n/5 + self.trailingZeroes(n/5)
       

循环

class Solution(object):
    def trailingZeroes(self, n):
        """
        :type n: int
        :rtype: int
        """
        #A very smart question, 2 is always ample, so we only need to care about the factor of 5
        res = 0
        while n > 0:
            n /= 5
            res += n
        return res