method recursion
method recursion
concept
Method recursion: Program structure, which is the basis in the data structure and algorithm section.
Data Structure: How data is stored.
Algorithms: How data is processed.
in conclusion
1. In what scenarios can recursion be used?
1. A large problem can be divided into several sub-problems;
2. The original problem and the sub-problem have the same solution except for the different data scales;
3. There is a recursive termination condition.
2. What is method recursion?
The process of a method calling itself is called method recursion (remember the semantics of the method!!! Be sure not to think about internal implementation)
Example: Write a recursive function to implement the factorial of 5
public class Recursion {
public static void main(String[] args) {
int num = 5;
System.out.println(factor(5));
}
//写一个递归函数实现5！
//一定注意方法的语义（这个方法能干嘛）
//factor(num) 传入 num -> num!
//5! 5 * factor (4)
public static int factor(int num){
//先写终止条件
if (num == 1){
return num;
}
return num * factor(num - 1);
}
}
Example: Implementing 1+2+…+10 with recursion
public class Recursion {
public static void main(String[] args) {
int num = 10;
System.out.println(add(10));
}
public static int add(int num){
if (num == 1){
return 1;
}
return num + add(num - 1);
}
}
Example: Enter a non-negative integer and return the sum of the numbers that make it up
public class Recursion {
public static void main(String[] args) {
int num = 1729;
System.out.println(sum(num));
}
//输入一个非负整数，返回组成他的数字之和
public static int sum(int num){
//num只有个数
//终止条件
if (num < 10){
return num;
}
//num > 10
//能知道num的个位
return num % 10 + sum(num / 10);
}
}
Example: Print each digit of a number in sequence
public class Recursion {
public static void main(String[] args) {
int num = 1729;
print(num);
}
//按顺序打印数字的每一位
//传入一个num就能打印每一位的值
public static void print(int num){
if (num > 10){
//递的过程
//此时num还需要继续寻址最高位
print(num / 10);
}
//此时说明最高位已经找到
//%10是为了保留其他位
System.out.print(num % 10 + " ");
}
}
Example: frog jumping steps
Question: A frog can jump up one step or two steps at a time. How many ways can the frog jump up n steps?
public class Homework2 {
public static void main(String[] args) {
System.out.println(frogJump(5));
}
/*
例1：青蛙跳台阶问题-----斐波那契数的应用
题：一只青蛙一次可以跳上一级台阶，也可以跳上两级，求该青蛙跳上n级台阶一共有多少种跳法？
*/
public static int frogJump(int n){
//先找终止条件 共有一级台阶 n = 1种
// 二 2种
if (n == 1){
return 1;
}
if (n == 2){
return 2;
}
return frogJump(n-1) + frogJump(n-2);
}
}
Example: Tower of Hanoi problem
How to achieve finding the largest plate? Put n-1 disks on B. At this time, only the largest disk
A—>C is left on A, and then n-1 disks on B are sequentially ordered from B—>C
public class Homework2 {
static int count = 0;
public static void main(String[] args) {
int n = 3;
char A = 'A';
char B = 'B';
char C = 'C';
hanoTower(n,A,B,C);
System.out.println("一个搬了"+count+"次盘子");
}
/*
例2.汉诺塔问题
A 起始位置
C 目标位置
B 辅助塔 暂存小盘子
*/
public static void hanoTower(int nDisks,char A,char B,char C){
if (nDisks == 1) {
//一个盘子，直接从A--->C
move(1,A,C);
return;
}
//此时有多个盘子
//核心步骤：先把n-1个盘子从A放到B上
hanoTower(nDisks - 1, A, C, B);
//此时找到了最大盘子A-->C
move(nDisks,A,C);
//再把B上的盘子依次从B——>C
hanoTower(nDisks-1, B, A, C);
}
/*
* 将标号为 n 的盘子从sourceTower(原塔）到distTower(目标塔）
* */
public static void move(int n,char sourceTower,char distTower){
count ++;
System.out.println("标号为"+n+"盘子从"+sourceTower+"--->"+distTower);
}
}
