Question

我有一个Doubly Linked List和I方法，它应该删除一个给定的节点。当我最初开发代码时，它最初与我的3个测试节点一起使用但是当我向Doubly Linked List添加更多节点时它停止工作，现在我得到一个NullReferenceException：“对象引用未设置为对象的实例”我知道我可以使用通用链表但我正在尝试学习双链表如何工作。

我将在我的代码中标记行，我在行的开头用“**”获取此异常（它位于Doubly Linked List类的RemoveNode方法中）。

这是我的代码：

Doubly Linked List类：

class DoublyLinkedList
{
    public Node head, current;

    public void AddNode(object n) // add a new node 
    {
        if (head == null)
        {
            head = new Node(n); //head is pointed to the 1st node in list 
            current = head;
        }
        else
        {
            while (current.next != null)
            {
                current = current.next;
            }

            current.next = new Node(n, current); //current is pointed to the newly added node 
        }
    }

    public Node FindNode(object n) //Find a given node in the DLL
    {
        current = head;
        while ((current != null) && (current.data != n))
            current = current.next;

        if (current == null)
            return null;
        else
            return current;                    
    }


    public string RemoveNode(object n)//remove nodes
    {
        String Output = "";

        if (head == null)
        {
            Output += "\r\nLink list is empty";
        }
        else
        {
            current = FindNode(n);

            if (current == null)
            {
                Output += "Node not found in Doubly Linked List\r\n";
            }
            else
            {                   
                ****current.next.prev = current.prev;**         
                current.prev.next = current.next;
                current.prev = null;
                current.next = null;
                Output += "Node removed from Doubly Linked List\r\n";
            }
        }
        return Output; 
    }

    public String PrintNode() // print nodes 
    {
        String Output = "";

        Node printNode = head;
        if (printNode != null)
        {
            while (printNode != null)
            {
                Output += printNode.data.ToString() + "\r\n";
                printNode = printNode.next;
            }
        }
        else
        {
            Output += "No items in Doubly Linked List";
        }
        return Output;
    }

}

节点类：

class Node
{
    public Node prev, next; // to store the links 
    public object data; // to store the data in a node 

    public Node()
    {
        this.prev = null;
        this.next = null;
    }

    public Node(object data)
    {
        this.data = data;
        this.prev = null;
        this.next = null;
    }

    public Node(object data, Node prev)
    {
        this.data = data;
        this.prev = prev;
        this.next = null;
    }

    public Node(object data, Node prev, Node next)
    {
        this.data = data;
        this.prev = prev;
        this.next = next;
    } 

}

Answer 1

我得到了它的工作。虽然我略微改变了你的逻辑。我使用current字段标记尾部，并使用单独的变量进行搜索。现在，我用整数值测试它，导致问题的是你比较值的行

(current.data != n)

你可以得到那个10！= 10，因为这些值被装箱并且引用不同。只需使用current.data.Equals(n)代替。 Equals()是一个虚拟方法，它向下冒泡到真实对象并比较数据而不是引用。

此外，您的删除逻辑必须更复杂。您还可以通过删除所有必要的if / else语句来使其更具可读性。

public void AddNode(object n) // add a new node 
{
    if (head == null)
    {
        head = new Node(n); //head is pointed to the 1st node in list 
        current = head;
    }
    else
    {
        var newNode = new Node(n, current);
        current.next = newNode; 
        current = newNode; //current is pointed to the newly added node 
    }
}

public Node FindNode(object n) //Find a given node in the DLL
{
    var index = head;
    while (index != null)
    {
        if (index.data.Equals(n))
            break;

        index = index.next;
    }

    return index ?? null;
}

public string RemoveNode(object n) //remove node
{
    if (head == null)
        return "\r\nLink list is empty";

    var node = FindNode(n);

    if (node == null)
        return "Node not found in Doubly Linked List\r\n";        

    if (node != head)
        node.prev.next = node.next;

    if (node.next != null)
        node.next.prev = node.prev;

    return "Node removed from Doubly Linked List\r\n";
}

Answer 2

您是否尝试从仅包含一个元素（= head）的列表中删除对象？你在那里得到一个错误，检查你的删除例程。您正在使用FindNode（然后将返回您的头节点）收到一个元素，之后您正在调用current.prev.next = current.next;，这将失败，因为您的headnode既没有previous也没有{ {1}}节点。

我认为你的删除功能应该类似于：

next

从双向链接列表中删除节点时遇到问题

2 个答案: