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我是java的初学者，我目前在工作节点。我想知道是否有一种方法可以显示列表的内容而不必使用.getNext方法，因为一旦我使用它，它就会删除节点上的元素，从字面上删除顶部的节点。我在这段代码中试图做的是使用输入将两个String元素存储在新的两个节点中，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方法来检查列表。 请注意，在Book的某些奇怪的原因类 T不显示除了类和实施的类，如果我把<>围绕它。显示LinkedStack的内容而不删除它们
继承人的代码：
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，其中包含的方法
公共类图书实现MyInterface的
{
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）为了更好地提供帮助，请发布[SSCCE]（http://sscce.org/）。 2）对代码块使用一致的逻辑缩进。代码的缩进旨在帮助人们理解程序流程！ –
谢谢先生。我会记住这一点。 –