java优秀算法河内之塔_河内塔的Java程序

java优秀算法河内之塔

Tower of Hanoi is a mathematical puzzle where we have three rods and n disks. The objective of the puzzle is to move all disks from source rod to destination rod using the third rod (say auxiliary). The rules are:

河内塔是一个数学难题，其中我们有三个杆和n个盘。 难题的目的是使用第三个杆(例如辅助杆)将所有磁盘从源杆移动到目标杆。 规则是：

1. Only one disk can be moved at a time.

   一次只能移动一个磁盘。

2. A disk can be moved only if it is on the top of a rod.

   仅当磁盘位于杆的顶部时才能移动磁盘。

3. No disk can be placed on the top of a smaller disk.

   不能将磁盘放置在较小磁盘的顶部。

Print the steps required to move n disks from source rod to destination rod. Source Rod is named as 'a', auxiliary rod as 'b' and destination rod as 'c'.

打印将n个磁盘从源棒移到目标棒所需的步骤。 源杆称为'a' ，辅助杆称为'b' ，目的杆称为'c' 。

Input format: Integer n

输入格式：整数n

Output format: Steps in different lines (in one line print source and destination rod name separated by space).

输出格式：不同行中的步骤(在一行中，打印源和目标杆名称用空格分隔)。

Example:

例：

    Sample Input:
    2

    Sample Output:
    a b
    a c
    b c

Explanation:

说明：

This is one of the famous problems on recursion. To solve this, let's assume the steps taken for 2 disks. Let’s assume Rod A, B, and C and we have to shift the disks from A to B. First, we shift the smaller disk to C, then we shift the larger disk to B, and at last, put the smaller disk from C to B.

这是递归的著名问题之一。 为了解决这个问题，我们假设对2个磁盘采取了步骤。 假设杆A，B和C，我们必须将磁盘从A移到B。首先，将较小的磁盘移至C，然后将较大的磁盘移至B，最后，将较小的磁盘从C移入到B。

Therefore, for N disks, lets recursion shifts N-1 disks to C, and we will shift the last disk to B and again let recursion shifts rest of the disk to C. Using this, we will be able to solve the problem.

因此，对于N个磁盘，让递归将N-1个磁盘移至C，然后将最后一个磁盘移至B，再让递归将其余磁盘移至C。使用此方法，我们可以解决问题。

Algorithm:

算法：

Declare a recursive function towerOfHanoi with parameters (int disk, char source, char auxiliary, char destination)

声明带有参数(int磁盘，char源，char辅助，char目标)的递归函数towerOfHanoi

• STEP 1: Base Case: If(disk == 0) return;

  步骤1：基本情况： If(disk == 0)返回；

• STEP 2: Base Case: If(disk == 1) Print Source to Destination

  步骤2：基本情况： If(disk == 1)打印源到目标

• STEP 3: Recursive Case: towerOfHanoi(disk -1, source, destination, auxiliary)

  步骤3：递归案例： towerOfHanoi(磁盘-1，源，目标，辅助)

• STEP 4: Print Source to Destination

  步骤4：将源打印到目标

• STEP 5: towerOfHanoi(disk -1, auxiliary, source,destination)

  步骤5： TowerOfHanoi(磁盘-1，辅助，源，目标)

Example:

例：

    Input : 3

    Disk 1 moved from A to C
    Disk 2 moved from A to B
    Disk 1 moved from C to B
    Disk 3 moved from A to C
    Disk 1 moved from B to A
    Disk 2 moved from B to C
    Disk 1 moved from A to C

Program:

程序：

import java.util.Scanner;

public class Main {
     
	//Recursive Function
	public static void towerOfHanoi(int disks, char source, char auxiliary, char destination) {
      
		// Write your code here
		if(disks==0){
        //Base Case 1
			return;
		}
		if(disks==1){
        //Base Case 2
			System.out.println(source +" " + destination);
			return;
		}
		else{
     
			//Shifting d-1 disk from A to C
			towerOfHanoi(disks-1,source,destination,auxiliary);   
			System.out.println(source + " " + destination);
			//Shifting d-1 disk from c to B
			towerOfHanoi(disks-1,auxiliary,source,destination);   
		}
	}

	public static void main(String[] args) {
     
		int disk;
		Scanner s = new Scanner(System.in);
		
		System.out.print("Print No of Disks: ");
		disk = s.nextInt();
		
		towerOfHanoi(disk, 'A', 'C', 'B');
	}
}

Output

输出量

Print No of Disks: 3
A B
A C
B C
A B
C A
C B
A B

翻译自: https://www.includehelp.com/java-programs/tower-of-hanoi.aspx

java优秀算法河内之塔