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

Note: Your solution should be in logarithmic time complexity.

 此题是求阶乘后面零的个数。

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

每天一道LeetCode--172. Factorial Trailing Zeroes