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

Note: Your solution should be in logarithmic time complexity.

一刷
題解：
求n!尾部有多少個0。其實主要就是看有多少個5，有多少個5就有多少個0
count = n / 5 + n / 25 + n / 125 + ....
如果有25，會產(chǎn)生2個0， 125會產(chǎn)生3個0

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