Leetcode 172 Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

 

A number multiplied by 10 will have a trailing 0 added to it. So we only need to find out how many 10's will appear in the expression of the factorial. Since 10 = 2 * 5 and there are a bunch more 2's (each even number will contribute at least one 2), we only need to count the number of 5's.

Well, how to count them all? If you try some examples, you may finally get the result, which is n / 5 + n / 25 + n / 125 + ....

class Solution:

    # @param {integer} n

    # @return {integer}

    def trailingZeroes(self, n):

        ans = 0

        while n:

            ans += n / 5

            n /= 5

        return ans

递归简化版本

class Solution:

    # @param {integer} n

    # @return {integer}

    def trailingZeroes(self, n):

        return n and n/5 + self.trailingZeroes(n/5)