【问题标题】:Towers Of Hanoi with Stacks Implementation and Recursion (Java)具有堆栈实现和递归的 Hanoi Towers (Java)
【发布时间】:2018-03-26 08:10:11
【问题描述】:

我正在尝试通过递归解决河内塔问题,同时使用堆栈。输出应如下所示,使用 n # 个大小不断增加的磁盘。磁盘必须从 Pillar1 移动到 Pillar3,一次一个:

// assuming n = 3;

Pillar1: 3 2 1

Pillar2:

Pillar3: 

// "\n"

Pillar1: 3 2 

Pillar2:

Pillar3: 1

// "\n"

Pillar1: 3 

Pillar2: 2

Pillar3: 1


// "\n"

Pillar1: 3 

Pillar2: 2 1

Pillar3: 


// "\n"

Pillar1: 

Pillar2: 2 1

Pillar3: 3


// "\n"

Pillar1: 1

Pillar2: 2

Pillar3: 3

// "\n"

Pillar1: 1

Pillar2: 

Pillar3: 3 2


// "\n"

Pillar1: 

Pillar2: 

Pillar3: 3 2 1

我的代码在下面,我很难输出磁盘 > 1:

import java.util.*;
class TowersOfHanoiThree
{
   public static Stack<Integer>[] tower = new Stack[4];
   public static int temp;

   public static void TowersOfHanoiThree(int numDisk)
   {
      //adding disk to stack
      temp = numDisk;
      tower = new Stack[4];

      for(int a = 0; a <= 3; a++)
      {
         tower[a] = new Stack<Integer>();
      }

     for (int i = numDisk; i > 0; i--)
     {
        tower[1].push(numDisk);
        show();
     }
     solver(numDisk, 1, 3, 2);
 }

public static void show()
{
   //System.out.println("Pillar1: ");
   //System.out.println("Pillar2: ");
   //System.out.println("Pillar3: ");

   String Pillar1 = "Pillar1: ";
   String Pillar2 = "Pillar2: ";
   String Pillar3 = "Pillar3: ";

   for(int x = temp -1 ; x >= 0 ; x--)
   {
      String emptStr1 = "";
      String emptStr2 = "";
      String emptStr3 = "";

     try
     {
        emptStr1 = String.valueOf(tower[1].get(x));
     }
     catch(Exception e)
     {
     }

     try
     {
        emptStr2 = String.valueOf(tower[2].get(x));
     }
     catch(Exception e)
     {
     }

     try
     {
        emptStr3 = String.valueOf(tower[3].get(x));
     }
     catch(Exception e)
     {
     }
     System.out.print(Pillar1+emptStr1+"\n");
     System.out.print(Pillar2+emptStr2+"\n");
     System.out.print(Pillar3+emptStr3+"\n");
     System.out.print("\n");
  }
}//end show

public static void solver(int numDisk, int start, int middle, int end) 
{
   if(numDisk > 0) 
   {
      try
      {
         //sorting disks
         solver(numDisk - 1, start, end, middle);
         int dis = tower[start].pop(); //move disk top-most disk of start
         tower[middle].push(dis);
         show();
         solver(numDisk - 1, middle, start, end);
      }
      catch(Exception e)
      {
      }
   }
}

public static void main(String args[])
{
    tower[1] = new Stack<Integer>();
    tower[2] = new Stack<Integer>();
    tower[3] = new Stack<Integer>();

    TowersOfHanoiThree(2);
}
}

【问题讨论】:

    标签: java algorithm recursion stack towers-of-hanoi


    【解决方案1】:

    使用以下实现应该可以为您提供所需的结果。我还放置了一些经过修改的内联 cmets

    public static void TowersOfHanoiThree(int numDisk)
    {
      //adding disk to stack
      temp = numDisk;
      tower = new Stack[4];
    
      for(int a = 0; a <= 3; a++)
      {
         tower[a] = new Stack<Integer>();
      }
    
     for (int i = numDisk; i > 0; i--)
     {
        tower[1].push(i); // to show "1 2 3" i changed the value which was pushed in the stack
    // from tower[1].push(numDisk) to tower[1].push(i)
     }
     show(); //In your example this method call was placed inside the for loop.
    //Moving it outside will show the stacks only after the values are added
     solver(numDisk, 1, 3, 2);
    }
    
     public static void show() {
        // System.out.println("Pillar1: ");
        // System.out.println("Pillar2: ");
        // System.out.println("Pillar3: ");
    
        String Pillar1 = "Pillar1: ";
        String Pillar2 = "Pillar2: ";
        String Pillar3 = "Pillar3: ";
    
        String emptStr1 = "";
        String emptStr2 = "";
        String emptStr3 = "";
    //the empStr variable are moved outside the loop because only after the 
    //loop has finished we know what values are in each pillar
        for (int x = 0; x <= temp - 1; x++) {
    
            try {
    //here we just append the values from the pillar to string empStr1
                emptStr1 += String.valueOf(tower[1].get(x)) + " ";
            } catch (Exception e) {
            }
    
            try {
                emptStr2 += String.valueOf(tower[2].get(x)) + " ";
            } catch (Exception e) {
            }
    
            try {
                emptStr3 += String.valueOf(tower[3].get(x)) + " ";
            } catch (Exception e) {
            }
        }
    //finally when the loop is done we have the results
        System.out.print(Pillar1 + emptStr1 + "\n");
        System.out.print(Pillar2 + emptStr2 + "\n");
        System.out.print(Pillar3 + emptStr3 + "\n");
        System.out.print("\n");
    }// end show
    

    【讨论】:

      【解决方案2】:

      您需要解决一些问题。我会尽力解释它们并提供可能的解决方案。

      首先,您总是在 TowersOfHanoiThree(int) 方法中将 numDisks 推送到您的初始堆栈(支柱)上。 违规部分:

      for (int i = numDisk; i > 0; i--) {
         tower[1].push(numDisk); // should be tower[1].push(i);
         show();
      }
      

      修复:

      for (int i = numDisk; i > 0; i--) {
         tower[1].push(i); // fixed
         show();
      }
      

      其次,您的 show() 方法不正确,原因有很多:

      1. 您应该从索引 0 而不是最后一个索引 (temp - 1) 开始迭代数组 tower 中的每个堆栈,因为您希望从底部到顶部显示磁盘。
      2. 您应该在for 循环之前声明您的字符串emptStr1emptStr2emptStr3。目前,您在每次迭代时将它们重置为空字符串。
      3. 您需要将磁盘连接到字符串emptStr1emptStr2emptStr3,而不是通过重新分配来覆盖它们。
      4. 您应该在 for 循环之后输出您的字符串 emptStr1emptStr2emptStr3

      我建议完全重写show(),例如:

      public static void show() {
          for (int i = 1; i < tower.length; i++) { // iterate over your array of stacks
              String pillar = "Pillar " + i + ":"; // e.g. "Pillar 1: "
              for (int disk : tower[i]) { // iterate over the stack located at index i
                  pillar += " " + disk; // concatenate the current disk to the String
              }
              System.out.println(pillar); // Display this pillar
          }
          System.out.println(); // Line break to seperate the show() calls
      }
      

      终于solver(int, int, int, int)中的第二次递归调用不正确。

      违规行:

      solver(numDisk - 1, middle, start, end);
      

      这应该改为:

      solver(numDisk - 1, end, middle, start);
      

      另外,虽然不是失败的原因,但您的数组 tower 的容量为 4,但应该是 3。您目前忽略索引 0。我想您是故意这样做的因为您想将支柱显示为 1、2 和 3?无论如何,这可能会令人困惑。相反,当您显示您的支柱信息时,只需将 1 添加到您的索引。

      【讨论】:

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