两个双向链表串联JAVA

Question

我在实现两个双向链表的连接时遇到了麻烦。这是我的 concat 方法。看起来不错，但结果我得到了第一个列表的尾部元素，以及第二个列表的所有元素。

    public Node<?> concat(Node<?> head1, Node<?> head2)
{
    if(head1 == null)
    {
        return head2;
    }
    if(head2 == null)
    {
        return head1;
    }

    Node<?> n = head1;

    while(n.getNext() != null)
    {
        n = n.getNext();
    }
    n.setNext(head2);

    return head1;
}

编辑：DoubleLinkedList 类：

public class DoubleLinkedList<E>
{
protected int size;
protected Node<?> head, tail;

public DoubleLinkedList()
{
    size = 0;
    clear();
}

public void clear()
{
    head = null;
    tail = null;
}

public int size()
{
    return size;
}

public boolean isEmpty()
{
    return head == null;
}

public Node<?> getHead()
{
    return head;
}

public Node<?> getTail()
{
    return tail;
}

public void add(E value)
{
    Node<E> node = new Node<E>(value);
    if(isEmpty())
    {
        head = node;
        tail = node;
    }
    else
    {
        tail.setNext(node);
        node.setPrevious(tail);
        tail = node;
    }
    size++;
}

public int indexOf(E value)
{
    Node<?> find = head;
    for(int i=0;i<size;i++)
    {
        E n = (E}find.getValue();
        if(n.equals(value))
        {
            return i;
        }
        find = find.getNext();
    }

    return -1;
}


public Node<?> concat(Node<?> head1, Node<?> head2)
{
    if(head1 == null)
    {
        return head2;
    }
    if(head2 == null)
    {
        return head1;
    }

    Node<?> n = head1;

    while(n.getNext() != null)
    {
        n = n.getNext();
    }
    n.setNext(head2);
    head2.setPrevious(n);

    return n;
}


private static final class Node<E>
{
    private E value;
    private Node<?> next, previous;

    public Node(E value)
    {
        this(value, null, null);
    }

    public Node(E value, Node<?> n, Node<?> p)
    {
        this.value = value;
        this.next = n;
        this.previous = p;
    }

    public E getValue()
    {
        return value;
    }

    public void setNext(Node<?> n)
    {
        this.next = n;
    }

    public Node<?> getNext()
    {
        return next;
    }

    public void setPrevious(Node<?> p)
    {
        this.previous = p;
    }

    public Node<?> getPrevious()
    {
        return previous;
    }
}

}

Answer 1

不要返回head1，而是尝试返回n。