Moderate 阶层的结尾0个数 @CareerCup

思路和《经典面试题：100的阶乘有几个结尾零》是一样的，但代码更简洁了，尤其是第二种方法！

package Moderate;/** * Write an algorithm which computes the number of trailing zeros in n * factorial. *  * 译文： *  * 写一个算法计算n的阶乘末尾0的个数。 *  */public class S17_3 {public static int factorsOf5(int i) {int count = 0;while (i % 5 == 0) {count++;i /= 5;}return count;}// 第一种方法就是对阶层的所有因子都计算有几个5因子public static int countFactZeros(int num) {int count = 0;for (int i = 2; i <= num; i++) {count += factorsOf5(i);}return count;}// 第二种方法，更快地计算出总数public static int countFactZeros2(int num) {        int count = 0;        if (num < 0) {                System.out.println("Factorial is not defined for negative numbers");                return 0;        }        while(num > 0){        count += num / 5;        num /= 5;        }        return count;}public static int factorial(int num) {if (num == 1) {return 1;} else if (num > 1) {return num * factorial(num - 1);} else {return -1; // Error}}public static void main(String[] args) {for (int i = 1; i < 12; i++) {System.out.println(i + "! (or " + factorial(i) + ") has " + countFactZeros(i) + " zeros");System.out.println(i + "! (or " + factorial(i) + ") has " + countFactZeros2(i) + " zeros");}}}