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; } }