反转未按预期工作的链表

Question

我目前正在尝试反转一个链表，我有一个意外的问题，它不能正确打印。当我尝试访问链表的值时，我得到一个错误。如果有专家的眼光来看看我错过了什么，我会非常感激的。提前谢谢。

public class SinglyLinkedList implements Cloneable, Iterable, List {
    //---------------- nested Node class ----------------

    /**
     * Node of a singly linked list, which stores a reference to its
     * element and to the subsequent node in the list (or null if this
     * is the last node).
     */
    private static class Node {
        private E data;  // reference to the element stored at this node

        private Node nextNode; 

        
        //methods for accessing variables

        public Node getNextNode() { return nextNode; }

        public E getData() { return data; }

        // Modifier methods
        public void setNext(Node n) { nextNode = n; }
        public void setData(E n) { data = n; }
        public Node(E e, Node n) {
            nextNode = n;
            data = e;
        }

        // reference to the subsequent node in the list// TODO
    } //----------- end of nested Node class -----------

    // instance variables of the SinglyLinkedList
    private int size = 0; // number of nodes in the list
    private Node head = null; // head node of the list (or null if empty)

    public SinglyLinkedList() {
    }              // constructs an initially empty list

    // access methods

    /**
     * Returns the number of elements in the linked list.
     *
     * @return number of elements in the linked list
     */
public void addFirst(E e) {
        head = new Node(e, head); // create the new node and link new node
        size++;
    }
    /**
     * Produces a string representation of the contents of the list.
     * This exists for debugging purposes only.
     */
    public String toString() {
        StringBuilder temporaryString = new StringBuilder();
        temporaryString.append("[");
        for(Iterator iterator = iterator(); iterator.hasNext();){
            temporaryString.append(iterator.next()).append(", ");
        }
        temporaryString.deleteCharAt(temporaryString.length() - 1);
        temporaryString.deleteCharAt(temporaryString.length() - 1);
        temporaryString.append("]");
        return temporaryString.toString();
    }

    private class SinglyLinkedListIterator implements Iterator {
        private Node current;
        public SinglyLinkedListIterator() {
            current = head;
        }
        @Override
        public boolean hasNext() {
            return current != null;
        }

        @Override
        public E next() {
            if(!hasNext()) throw new RuntimeException("No such element");

            E res = current.getData();
            current = current.getNextNode();
            return res;
        }
    }

    public Iterator iterator() {
        return new SinglyLinkedListIterator();
    }

反向链表方法：

public Node reverseLinkedList(SinglyLinkedList sll) {
        Node previous = null;
        Node current = sll.head;
        while (current != null) {
            Node nextElement = current.getNextNode();
            current.setNext(previous);
            previous = current;
            current = nextElement;
        }
        return previous;
    }

主要方法：

public static void main(String[] args) {
        SinglyLinkedList sll2 = new SinglyLinkedList();
        sll2.addFirst(1);
        sll2.addFirst(2);
        sll2.addFirst(3);

System.out.println(sll2.toString());
        sll2.reverseLinkedList(sll2);
System.out.println(sll2.toString());
    }

输出：

[3, 2, 1]
//i should expect to get 1,2,3
[3]

Answer 1

就像你一样

变异

(“重新连接”)方法中给定的链表。

函数时，实际上并不是在生成新的链表。所以让它返回一些东西实际上是矛盾的。要么它应该返回一个全新的链表而不改变给定的链表，要么它应该改变给定的链表而不返回任何东西。

来自您的

我看到的代码实际上是你想要的

变异

给定的链表，因为您不使用返回值并根据

..。因此，将您的函数设置为

一，然后删除

语句。

现在的核心问题是您永远不会更改

成员:它仍然引用以前是列表中第一个节点的内容，但现在是最后一个节点。您应该将该节点分配给

这之前是尾节点。

因此，函数中的最后一条语句应该是：

sll.head = previous;