[Lintcode]2. Trailing Zeros

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于 2019-02-10 03:22:00 发布

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原文链接：http://www.cnblogs.com/siriusli/p/10358619.html

本文介绍了一种高效计算任意正整数阶乘尾部零的数量的算法，该算法利用了数学和位操作原理，避免了传统算法中复杂的算术运算，实现了O(logN)的时间复杂度。

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2. Trailing Zeros

本题难度: Easy
Topic: Math&Bit Manipulation

Description

Write an algorithm which computes the number of trailing zeros in n factorial.

Example
Example 1:
Input: 11
Output: 2

Explanation: 
11! = 39916800, so the output should be 2

Example 2:
Input: 5
Output: 1

Explanation: 
5! = 120, so the output should be 1.

Challenge
O(log N) time

我的代码

class Solution:
    """
    @param: n: An integer
    @return: An integer, denote the number of trailing zeros in n!
    """
    def trailingZeros(self, n):
        # write your code here, try to do it without arithmetic operators.
        count = 0
        f = 5
        while(f<n):
            count += n//f
            f *= 5
        return count