[LeetCode] 172. Factorial Trailing Zeroes_Easy tag: Math

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.

思路是因为Because all trailing 0 is from factors 5 * 2, 然后2这个因子总是足够的, 比如2, 4, 6, 每两个就有个2的因子, 然后我们只需要数5的个数即可, 另外要反复循环直到没有5为止.

比如1,2,3,4,5,6,7,8,9,10

发现 1-4: 0

5-9: 1

　　  10-14: 2

Code(O(lgn))

class Solution:
    def FactorialZeroTail(self, n):
        return 0 if n == 0 else n//5 + self.FactorialZeroTail(n//5)