java優秀算法河內之塔

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:

河內塔是一個數學難題，其中我們有三個桿和n個盤。 難題的目的是使用第三個桿(例如輔助桿)將所有磁盤從源桿移動到目標桿。 規則是：

Only one disk can be moved at a time.

一次只能移動一個磁盤。

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

僅當磁盤位于桿的頂部時才能移動磁盤。

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

不能將磁盤放置在較小磁盤的頂部。

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'.

打印將n個磁盤從源棒移到目標棒所需的步驟。 源桿稱為'a' ，輔助桿稱為'b' ，目的桿稱為'c' 。

Input format: Integer n

輸入格式：整數n

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

輸出格式：不同行中的步驟(在一行中，打印源和目標桿名稱用空格分隔)。

Example:

例：

Sample Input:2Sample Output:a ba cb c

Explanation:

說明：

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.

這是遞歸的著名問題之一。 為了解決這個問題，我們假設對2個磁盤采取了步驟。 假設桿A，B和C，我們必須將磁盤從A移到B。首先，將較小的磁盤移至C，然后將較大的磁盤移至B，最后，將較小的磁盤從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，對于N個磁盤，讓遞歸將N-1個磁盤移至C，然后將最后一個磁盤移至B，再讓遞歸將其余磁盤移至C。使用此方法，我們可以解決問題。

Algorithm:

算法：

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

聲明帶有參數(int磁盤，char源，char輔助，char目標)的遞歸函數towerOfHanoi

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

  步驟1：基本情況： If(disk == 0)返回；

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

  步驟2：基本情況： If(disk == 1)打印源到目標

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

  步驟3：遞歸案例： towerOfHanoi(磁盤-1，源，目標，輔助)

• STEP 4: Print Source to Destination

  步驟4：將源打印到目標

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

  步驟5： TowerOfHanoi(磁盤-1，輔助，源，目標)

Example:

例：

Input : 3Disk 1 moved from A to CDisk 2 moved from A to BDisk 1 moved from C to BDisk 3 moved from A to CDisk 1 moved from B to ADisk 2 moved from B to CDisk 1 moved from A to C .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}} .minHeight{min-height: 250px;}@media (min-width: 1025px){.minHeight{min-height: 90px;}}

Program:

程序：

import java.util.Scanner;public class Main {//Recursive Functionpublic static void towerOfHanoi(int disks, char source, char auxiliary, char destination) { // Write your code hereif(disks==0){ //Base Case 1return;}if(disks==1){ //Base Case 2System.out.println(source +" " + destination);return;}else{//Shifting d-1 disk from A to CtowerOfHanoi(disks-1,source,destination,auxiliary); System.out.println(source + " " + destination);//Shifting d-1 disk from c to BtowerOfHanoi(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

輸出量

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

翻譯自: https://www.includehelp.com/java-programs/tower-of-hanoi.aspx

java優秀算法河內之塔

總結

以上是生活随笔為你收集整理的java优秀算法河内之塔_河内塔的Java程序的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。