leetcode 172 Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

解决思路:
决定阶乘末尾零的个数其实是数列中5出现的次数,比如5的阶乘一个零。1024的阶乘末尾到底有几个零呢?

http://bbs.csdn.net/topics/380161955

这里写图片描述
代码如下:

int trailingZeroes(int n) 
{
    int total = 0;

    while(n>=5)
    {
        n = (n-(n%5))/5;
        total = total + n;
    }

    return total;

}

python 的解决方案:

class Solution:
    # @return an integer
    def trailingZeroes(self, n):
        factor, count = 5, 0

        while True:
            curCount = n // factor
            if not curCount:
                break

            count += curCount
            factor *= 5

        return count
posted @ 2017-11-17 22:28  wangyaning  阅读(...)  评论(...编辑  收藏