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)