Factorial Trailing Zeroes 续

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

Note: Your solution should be in logarithmic time complexity.

 

有了前面C里的仔细研究，python实现应该是手到擒来，突然这时候才注意到Note里已经标明了，时间复杂度在log

可见就算时间复杂度为o（n）都无法通过，更不用提前面还来两层循环遍历的

照葫芦画瓢，很容易通过，提交

class Solution:
    # @return an integer
    def trailingZeroes(self, n):
        j = 0
        while n > 0:
            j += n / 5
            n = n / 5
        return j

结果也通过了

502 / 502 test cases passed.
Status: Accepted
Runtime: 80 ms

还想依靠python的内置模块来试试，比如math

import math
class Solution:
    # @return an integer
    def trailingZeroes(self, n):
        return math.factorial(n)

可是提交之后，结果莫名其妙

Status: Output Limit Exceeded
Submitted: 0 minutes ago

红色的，应该是没通过，但不清楚他想要表达什么~！