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package com.lintcode;

/**

* 2. 尾部的零

* 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

*

* O(logN)的时间复杂度

*/

public class Solution2 {

/*

* @param n: An integer

* @return: An integer, denote the number of trailing zeros in n!

*/

public long trailingZeros(long n) {

// write your code here, try to do it without arithmetic operators.

long count = 0;

while (n > 0){

count += n / 5;

n /= 5;

}

return count;

}

}