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;
}