Factorial Trailing Zeroes 阶乘的后边有几个0

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

Note: Your solution should be in logarithmic time complexity.

//所有的因子里边含有2和5的个数，而2的个数明显会多余5的，所以只需要计算含有因子5为几个就是几个0，像25*4=100，所以25含有2个5(10含有一个2和一个5,)



public class Solution {

    public int trailingZeroes(int n) {

        int sum=0;

        while(n>0){

            sum+=n/5;

            n=n/5;

        }

        return sum;

    }

}