使用toString打印出链表

Question

好的，我正在尝试学习如何打印出链表。我有需要用于列表的所有方法，但是我不知道如何显示节点的值。现在，我的main方法中什么都没有，因为在尝试在main方法中调用非静态方法时，我不断出错。我有一个toString方法，用于显示列表的内容。我将如何调用此toString以显示每个节点的值？任何建议将不胜感激。

这是节点类：

public class LinkedListNode
{

    private int data;
    private LinkedListNode next;


    public LinkedListNode(int data)
    {
        this.data = data;
        this.next = null;
    }

    public int getData()
    {
        return data;
    }

    public void setData(int d)
    {
        data = d;
    }

    public LinkedListNode getNext()
    {
        return next;
    }

    public void setNext(LinkedListNode n)
    {
        next = n;
    }
}

这是LinkedList类，其中包含操作列表的main和方法：

public class LinkedList {

    public LinkedListNode head;

    public static void main(String[] args) {

    LinkedList l = new LinkedList();
    l.insertFront(0);
    System.out.println(l.toString());

    }

    public LinkedList() {
        this.head = null;
    }

    public int removeFront(){
        if(head == null){
            System.out.println("Error - Attempting to call removeFront() on empty list");
            return 0;
        }else{
            int temp = head.getData();
            head = head.getNext();  
            return temp;
        }

    }

    public void insertFront(int data){
        if(head == null){
            head = new LinkedListNode(data);
        }else{
            LinkedListNode newNode = new LinkedListNode(data);
            newNode.setNext(head);
            head = newNode;
        }       
    }

    public void insertBack(int data){
        if(head == null){
            head = new LinkedListNode(data);
        }else{
            LinkedListNode newNode = new LinkedListNode(data);
            LinkedListNode current = head;
            while(current.getNext() != null){
                current = current.getNext();
            }
            current.setNext(newNode);
        }       
    }

    public int removeBack(){
        if(head == null){
            System.out.println("Error - Attempting to call removeBack() on empty list");
            return 0;
        }else if (head.getNext() == null){
            int temp = head.getData();
            head = null;
            return temp;
        }else{

            LinkedListNode current = head;
            while(current.getNext().getNext() != null){
                current = current.getNext();
            }
            int temp = current.getNext().getData();
            current.setNext(null);
            return temp;
        }       
    }

    public String toString(){
        String retStr = "Contents:\n";

        LinkedListNode current = head;
        while(current != null){
            retStr += current.getData() + "\n";
            current = current.getNext();

        }

        return retStr;
    }

    public LinkedListNode getHead() {
        return head;
    }

    public void setHead(LinkedListNode head) {
        this.head = head;
    }
}

Answer 1

public static void main(String[] args) {

    LinkedList list = new LinkedList();
    list.insertFront(1);
    list.insertFront(2);
    list.insertFront(3);
    System.out.println(list.toString());
}

String toString() {
            String result = "";
            LinkedListNode current = head;
            while(current.getNext() != null){
                result += current.getData();
                if(current.getNext() != null){
                     result += ", ";
                }
                current = current.getNext();
            }
            return "List: " + result;
}