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：
Only one disk can be moved at a time.
一次只能移动一个磁盘。
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。
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
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