leetcode--Factorial Trailing Zeroes

题目：Factorial Trailing Zeroes

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

Note: Your solution should be in logarithmic time complexity.

计算n的阶乘的拖尾0的个数，即n中5的倍数的个数。

One：

public class Solution {    public int trailingZeroes(int n) {        if(n<0){            return 0;        }        int c = 0;        while(n/5!=0){            n/=5;            c+=n;        }        return c;    }}

Two：递归

public int trailingZeroes(int n) {    return n>=5 ? n/5 + trailingZeroes(n/5): 0;}