LeetCode 题解(204) : Factorial Trailing Zeroes

题目：

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

Note: Your solution should be in logarithmic time complexity.

题解：

0的个数由5的个数决定，但是25 = 5 * 5 等于两个5， 同理125， 625， 。。。。

递推公式为 r += n / 5, n /= 5

C++版：

class Solution {public:    int trailingZeroes(int n) {        int r = 0;        while(n > 0) {            int k = n / 5;            r += k;            n = k;        }        return r;    }};

Java版：

public class Solution {    public int trailingZeroes(int n) {        int r = 0;        while(n > 0) {            int k = n / 5;            r += k;            n = k;        }        return r;    }}

Python版：

class Solution(object):    def trailingZeroes(self, n):        """        :type n: int        :rtype: int        """        r = 0        while n > 0:            k = n / 5            r += k            n = k        return r