Java i'；我的链接列表有问题我可以显示列表的内容，但我找不到节点或列表的长度

对于linkNode长度，我建议使用递归函数，如下所示

public class IntNode {

    int data;
    IntNode link;

    IntNode() {

        this.link = null;

    }

    public IntNode(int initialData, IntNode initialLink) {

        data = initialData;
        link = initialLink;
    }

    public void addNode(int element) {
        link = new IntNode(element, link);
    }

    public int getData() {

        return data;
    }

    public IntNode getLink() {

        return link;
    }

    public void removeNodeAfter()

    {

        link = link.link;

    }

    public void setLink(IntNode newLink) {

        link = newLink;

    }

    void insertAtHead(int input) {
        input = data;
    }

    //this method doesnt work
    public int listLength(IntNode head) {
        IntNode cursor = 0;
        int ans;
        ans = 0;
        for (cursor = head; cursor != null; cursor = cursor.link)
        ans++;
        return ans;
    }

    //this method doesnt work either
    public boolean find(IntNode head, int searchKey) {

        if (head == null) {
            return false;
        }
        IntNode current = head;
        while (current != null) {
            if (current.data == searchKey) {
                return true;
            }
            current = current.link;
        }

        return false;

    }

import java.util.Scanner;

public class linkedList1 {

    public static void main(String[] args) {

        IntNode head;
        IntNode tail;

        head = new IntNode(10, null);
        head = new IntNode(20, head);
        head = new IntNode(25, head);
        head = new IntNode(30, head);

        while (head != null) {
            System.out.println(head.data);
            head = head.link;
        }

        IntNode node = new IntNode();

        System.out.println(node.listLength(null));

        if (node.find(head, 20)) System.out.println("found");

        else

          System.out.println("not found");

    }
}

public int listLength(IntNode head, int count){
    if(head.getLink() == null)
        return count;
    else
        return listLength(head.getLink(), count+1);
}

对于这样的函数，head为单个节点的调用listLength（head，1）将返回长度1，而具有两个节点的链表上的调用listLength（head，1）将返回2，以此类推

为了在这个链表中找到一个特定的值，我再次建议使用如下递归函数

public class IntNode {

    int data;
    IntNode link;

    IntNode() {

        this.link = null;

    }

    public IntNode(int initialData, IntNode initialLink) {

        data = initialData;
        link = initialLink;
    }

    public void addNode(int element) {
        link = new IntNode(element, link);
    }

    public int getData() {

        return data;
    }

    public IntNode getLink() {

        return link;
    }

    public void removeNodeAfter()

    {

        link = link.link;

    }

    public void setLink(IntNode newLink) {

        link = newLink;

    }

    void insertAtHead(int input) {
        input = data;
    }

    //this method doesnt work
    public int listLength(IntNode head) {
        IntNode cursor = 0;
        int ans;
        ans = 0;
        for (cursor = head; cursor != null; cursor = cursor.link)
        ans++;
        return ans;
    }

    //this method doesnt work either
    public boolean find(IntNode head, int searchKey) {

        if (head == null) {
            return false;
        }
        IntNode current = head;
        while (current != null) {
            if (current.data == searchKey) {
                return true;
            }
            current = current.link;
        }

        return false;

    }

import java.util.Scanner;

public class linkedList1 {

    public static void main(String[] args) {

        IntNode head;
        IntNode tail;

        head = new IntNode(10, null);
        head = new IntNode(20, head);
        head = new IntNode(25, head);
        head = new IntNode(30, head);

        while (head != null) {
            System.out.println(head.data);
            head = head.link;
        }

        IntNode node = new IntNode();

        System.out.println(node.listLength(null));

        if (node.find(head, 20)) System.out.println("found");

        else

          System.out.println("not found");

    }
}

public int listLength(IntNode head, int count){
    if(head.getLink() == null)
        return count;
    else
        return listLength(head.getLink(), count+1);
}

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希望这有帮助！另外，请将问题的格式设置得更好

您的链接列表代码没有问题。问题在于你的主要方法。我已经改正了。我把它作为一种练习留给你去找出你做错了什么

public boolean find(IntNode head, int searchKey){
    // Check if this is the correct node
    if(head.getData() == searchKey)
        return true;
    // Check if this is the last node
    else if(head.getLink() == null)
        return false;
    else
        return find(head.getLink(), searchKey);
}

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欢迎来到堆栈溢出！我想这段代码的95%与您的问题无关。请创建一个实例来演示您的问题。我理解@joe问题是什么？---------------配置：------30 25 20 10线程“main”java.lang.NullPointerException在IntNode.listLength（IntNode.java:57）在linkedList1.main（linkedList1.java:24）中出现异常过程已完成。我尝试使用您给我的代码，但我发现这个错误@Sishaar RaoYou需要提供更多上下文。抛出此错误的行是什么。也许调试是为了找出发生了什么，很可能您正在传入null，就像在System.out.println（node.listLength（null））中一样；我把它改为System.out.println（node.listLength（head，30））；然后确保你的头不是空的。调试并查看是否正确构造了链表。