172. Factorial Trailing Zeroes

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.

来自 <https://leetcode.com/problems/factorial-trailing-zeroes/description/>

 

思路1：耿直的想法，把阶乘算出来，然后计算末尾有多少个0 

class Solution(object):
    def trailingZeroes(self, n):
        """
        :type n: int
        :rtype: int
        """
        num = 1
        for i in range(1, n + 1):
            num = num * i
        num = str(num)[::-1]
        count = 0
        for i in num:
            if i == '0':
                count += 1
            else:
                break
        return count

 

思路2：问题的关键在于2和5的出现次数，因为只有2*5才能组成10，但2的出现概率是高于5的，所以只要计算5的出现次数就行了，如5！包含1个5，10！包含5和10这两个，25！包含5，10，15，10，25，其中25又可以分成5*5也就是6个。所以问题可以转换为求解包含5，5*5，5*5*5…的个数

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