设计一个算法,计算出n阶乘中尾部零的个数
样例
11! = 39916800,因此应该返回 2
class Solution {
/*
* param n: An desciption
* return: An integer, denote the number of trailing zeros in n!
*/
public long trailingZeros(long n) {
long count=0;
while(n!=0)
{
long temp=n/5;
count+=temp;
n=n/5;
}
return count;
}
};