面试总结--递归2

以下是在最近面试过程中碰到有关递归问题的题目：

/** * 汉诺塔问题 * * @param num * 盘子数 * @param a * 柱子a * @param b * 柱子b * @param c * 柱子c */public static void hannota(int num, char a, char b, char c) {if (num == 1) {move(1, a, c);return;} else {hannota(num - 1, a, c, b);move(num - 1, a, c);hannota(num - 1, b, a, c);}}static void move(int num, char source, char target) {System.out.println("moving " + num + " from " + source + " to "+ target);}

public class Recursion {public static void main(String[] args) {System.out.println("the last pepole's age is :" + getAge(10));System.out.println("the binary of the algorism is : " + getBinary(100));hannota(3, 'a', 'b', 'c');}/** * 第一个人的年龄是10，以后每个人的年龄都比前一个人的年龄大2岁，求第n个人的年龄 * * @param people * @return */public static int getAge(int people) {if (people == 1) {return 10;} else {return 2 + getAge(people - 1);}}/** * 求一个十进制数字的二进制形式是多少 * * @param num * @return */public static String getBinary(int num) {if (num < 2) {return num + "";} else {return getBinary(num / 2) + num % 2 + "";}}

/** * 写一个方法，要求参数int类型，如：传入一个12345，返回结果54321。用递归 */public class Reverse { public static void reverse(int number) { if (number < 10) { System.out.print(number); } else { int num = number % 10; System.out.print(num); reverse(number / 10); } } public static void main(String[] args) { reverse(123); }}