我的链表有问题，我可以显示列表的内容，但找不到节点或找不到列表的长度

Question

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");
    }
}

Answer 1

您的链接列表代码没有错。问题在于您的主要方法。我已经纠正了它。我把它作为练习，以弄清楚你做错了什么。

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 = null;
        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 ;
    }
}
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);
            IntNode cur = head;
            while(cur!= null){
                System.out.println(cur.data);
                cur= cur.link;
            }
            IntNode node = new IntNode();
            System.out.println(node.listLength(head));
            if(node.find(head, 20))
                System.out.println("found");
            else
                System.out.println("not found");

    }
}

Answer 2

对于LinkNode长度，我建议沿以下

递归函数

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，呼叫列表listlength（head，1），其中头是一个节点将返回长度1，而call listLength（head，1）在带有两个节点的链接列表上返回2，将返回2，等等。

对于此链接列表中的特定值，我再次建议沿以下

提出递归函数

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);
}

希望这会有所帮助！P.S.请更好地格式化您的问题