java - 删除链表中的节点后，打印节点列表显示删除的节点

Question

在下面的代码中，即使在删除节点 (20) 之后，如果我尝试通过将已删除的节点伪装成 head 来打印所有节点，它也会打印所有节点以及已删除的节点。有人可以解释这种行为以及 Java 中的垃圾收集吗？即使删除的节点 (20) 没有下一个元素，它如何能够迭代所有节点？

节点:

class Node{

    int nodeint;
    Node next;

    public Node(int nodeint){
        this.nodeint = nodeint;

    }

}

链表:

public class linkedlist{

    Node head;
    //Node next;
    public linkedlist(Node obj){

        this.head = obj;
    }

    public Node addnodes(int news){
        Node newlink = new Node(news);
        newlink.next = head;
        head = newlink;
        return head;
    }

    public void printAllNodes(Node obj){
        Node current  = obj;
        while(current!=null){
            System.out.println(current.nodeint);
            current = current.next;

        }

    }

    public Node remove(){

        Node temp = head;
        head = head.next;
        return temp;
    }

    public void printHead(){

        System.out.println("This is the present head node"+head.nodeint);

    }

    public static void main(String [] args){    

        Node obj1 = new Node(2);
        Node obj2 = new Node(3);
        Node obj3 = new Node(4);
        obj1.next  = obj2;
        obj2.next = obj3;
        obj3.next = null;
        linkedlist newobj = new linkedlist(obj1);
        Node obj = null;
        obj = newobj.addnodes(5);
        obj =newobj.addnodes(20);
        //System.out.println(obj.nodeint);
        newobj.printAllNodes(obj);
        obj = newobj.remove();
        System.out.println("A node is deleted");
        newobj.printAllNodes(obj);
        newobj.printHead();

    }
}

这段代码的输出:

20

5

2

3

4

A node is deleted

20

5

2

3

4

This is the present head node: 5

Answer 1

节点 20 仍然引用下一个节点，即本例中的节点 5。它与垃圾收集无关。如果您想要这种行为，请在您的remove 方法中设置 temp.next == null。