显示LinkedStack的内容而不删除它们

时间:2013-04-18 00:52:38

标签: java

我是java的初学者,我目前正在使用Nodes。我想知道是否有一种方法来显示列表的内容而不必使用.getNext方法,因为一旦我使用它,它就会删除Node上的元素,从根本上删除顶部的Node。我在这段代码中尝试做的是使用输入在新的两个节点中存储两个String元素,然后使用方法 proveTitle 来证明这些元素在列表中。一旦我这样做,我确保元素仍然完好无损,并使用toString方法检查列表。 请注意,在Book类中有一些奇怪的原因  如果我把<>放在了类和实现的类之外,T不显示围绕它。

下面是代码:

myNode类:

public class myNode<T> 
{

private T data;
private myNode next;

public myNode(T _data)
{
data = _data;
}
public myNode(T _data, myNode _next)
{
data = _data;
next = _next;
}

public T getData()
{
return data;    
}

public void setData(T _data)
{
data = _data;    
}

public myNode getNext()
{

return next;    
}

public void setNext(myNode _next)
{
 next = _next;   
}

}

接口类:

public interface myInterface<T> 
{
   public void pushTitle(T data);
   public T pop();
   public T peek();
   public String toString();
   public boolean isEmpty();
   public int size();
   public myNode getNode();
}

Book类,包含方法

public class Book实现myInterface

{
 private int count;
 private T author;
 private T title;
 private int stock;

 private myNode<T> top;

 public Book()
 {
     count = 0;
     top = null;
 }

@Override
 public myNode getNode()
 {
 return top;
 }


 @Override
 public void pushTitle(T title)
 {
 myNode<T> current = new myNode<>(title, top);
 current.setNext(top);
 top = current;
 count++;
 }

 public void proveTitle(T title)
 {
  T result;

  myNode<T> current = top;   

  if(title.equals(current.getData()))
  {
  result = current.getData();
  System.out.println("The title " + "'" + result + "'" + " exist."); 
  top = top.getNext();



 }
 }

 @Override
 public T pop()
 {
 T result;
 if(count == 0 || top == null )
 {
     System.out.println("List is empty");
 }
 System.out.println("The element on top is:"  +  top.getData());
 result = top.getData();
 top = top.getNext();

 count--;
 return result;

 }
 @Override

 public T peek()
 {
     System.out.println("Element on top is: " + top.getData());
     return top.getData();
 }
 @Override
 public boolean isEmpty()
 {
     if(top == null)
     {
     System.out.println("The list is empty");
     }
     else
     {
       System.out.println("The list is not empty." + "It has" + count + "elements");    
     }

  return top == null;   
 }
 @Override
 public int size()
 {
     System.out.println("The size of the list is" + count);
 return count;    
 }
 @Override
 public String toString()
 {
  String result = "";
  myNode current = top;
  System.out.println("Top");
  while(current != null)
  {
  result += ("[" + current.getData() + "]\n");
  current = current.getNext();
  }
  return result + "Bottom";
 }

}

主要类:

package node;
import java.util.Scanner;

public class myDriver 
{
    public static void main(String[]args)
    {
    Scanner input = new Scanner(System.in);  

    Book<String> title = new Book<>();



    myNode<String> current;
    current = title.getNode();
    String push;
    String push2;

    System.out.println("Enter title of book 1");
    push = input.nextLine();
    title.pushTitle(push);

    System.out.println("Enter title of book 2");
    push2 = input.nextLine();
    title.pushTitle(push2);

    title.proveTitle(push);
    title.proveTitle(push2);


    System.out.println(title.toString());






}
}

输出:

run:
Enter title of book 1
Tiger
Enter title of book 2
crossed
The title 'crossed' exist.
Top
[Tiger]
Bottom
BUILD SUCCESSFUL (total time: 7 seconds)

1 个答案:

答案 0 :(得分:0)

您似乎尝试使用List来实施Stack。这些是非常不同的数据结构。在List中,您要做的事情非常简单。在Stack中,它要求您有第二个堆栈。当您从当前堆栈弹出每个对象时,将其推送到第二个堆栈。

如果订单无关紧要,您可以执行此操作一次并切换到使用第二个堆栈。如果订单很重要,您将必须撤消该流程以恢复原始堆栈。

另请注意,更好的方法可能是使用List