Question

我正在创建一个从链表中删除节点的函数，但它给了我一个NullPointerException。我试图检查下一个是否为null，但现在它给了我那个错误。

删除功能：

 private boolean remove(Node aNode)
    {
        Node prevNode, nextNode;
        prevNode = this.getPrevious(aNode);
        if(aNode.getNext()==null){ // NullPointerException
            return false;
        }
        else{
            nextNode = aNode.getNext();
            prevNode.setNext(nextNode);
        }

        return false;
    }

节点类：

public class Node
{
    ///////////////////////////////////
    //           Properties          //
    ///////////////////////////////////
    private Object myData;
    private Node myNext;

    ///////////////////////////////////
    //             Methods           //
    ///////////////////////////////////

    /**
     *  Default constructor for a node with null
     *  data and pointer to a next node
     */
    public Node()
    {
        myData = null;
        myNext = null;
    }

    /**
     *  Constructor for a node with some object for
     *  its data and null for a pointer to a next node
     *
     *  <pre>
     *  pre:  a null node
     *  post: a node with some object for its data and
     *        null for a pointer to a next node
     *  </pre>
     *
     *  @param datum an object for the node's data
     */
    public Node(Object datum)
    {
        myData = datum;
        myNext = null;
    }

    /**
     *  Constructor for a node with some object for 
     *  its data and a pointer to another node
     *
     *  <pre>
     *  pre:  a null node
     *  post: a node with some object for its data and
     *        a pointer to a next node
     *  </pre>
     *
     *  @param datum an object for the node's data
     *  @param next the node that this node points to
     */
    public Node(Object datum, Node next)
    {
        myData = datum;
        myNext = next;
    }

    // Accessor methods
    public void setData(Object datum)
    {
        myData = datum;
    }

    public Object getData()
    {
        return myData;
    }

    public void setNext(Node next)
    {
        myNext = next;
    }

    public Node getNext()
    {
        return myNext;
    }
}

以下是完整Linked List类的主要部分

public static void main(String[] args)
    {
        LinkedList linkedList;
        Node testNode1, testNode2, testNode10, foundNode;
        boolean success;

        linkedList = new LinkedList();

        // Test "inList()" method
        testNode1 = new Node(new Integer(1));
        testNode2 = new Node(new Integer(2));
        testNode10 = new Node(new Integer(10));

       // System.out.println("In List = "+linkedList.inList(null));
        linkedList.printList();
        foundNode = linkedList.findNode(new Integer(2));
        System.out.println("Found node "+foundNode);
        success = linkedList.remove(null);
        System.out.println("Success = "+success);
        success = linkedList.remove(testNode1);
        System.out.println("Success = "+success);
        linkedList.addFirst(testNode1);
        success = linkedList.remove(testNode1);
        System.out.println("Success = "+success);
        linkedList.printList();
       // System.out.println("In List = "+linkedList.inList(null));
       // System.out.println("In List = "+linkedList.inList(testNode1));
       // System.out.println("In List = "+linkedList.inList(testNode2));

        // Test "addLast()" and "addFirst()" methods
        linkedList.addLast(new Node(new Integer(1)));
        linkedList.addLast(new Node(new Integer(2)));
        linkedList.addLast(new Node(new Integer(3)));
        linkedList.addLast(testNode10);
        foundNode = linkedList.findNode(new Integer(2));
        System.out.println("Found node "+foundNode.toString());
        linkedList.printList();

        Node testNode;
        testNode = linkedList.getPrevious(foundNode);
        System.out.println(testNode.getData());
        System.exit(0);

        success = linkedList.insertBefore("H", testNode1);
        System.out.println("Success = "+success);
        linkedList.printList();
        linkedList.addFirst(new Node(new Integer(1)));
        linkedList.addFirst(new Node(new Integer(2)));
        linkedList.addFirst(new Node(new Integer(3)));
        linkedList.printList();
        success = linkedList.insertBefore("A", testNode10);
        System.out.println("Success = "+success);
        linkedList.printList();

        // Test "remove()"
        success = linkedList.remove(testNode1);
        System.out.println("Success = "+success);
        success = linkedList.remove(testNode2);
        System.out.println("Success = "+success);
        success = linkedList.remove(testNode10);
        System.out.println("Success = "+success);
        linkedList.printList();
    }

}

Answer 1

您获得该异常是因为aNode为null并且您尝试调用null对象的getNext()方法，这意味着您在某个时候调用{{1} }。由于您没有告诉我们您拨打remove(null)的位置，因此无法确定，但您需要确保不会发生这种情况，或明确检查remove()是aNode在尝试调用方法之前。

如果您期待 null不是aNode但是它是，您应该仔细检查您的代码，以确保您实际上正确实施所有内容，因为这是一个很好的迹象，表明算法中的其他地方出了问题。

更新（使用新代码查看已编辑的问题）：您有：

null

这是你问题的根源;我的上述答案涵盖了修复例外的选项。

将来您需要检查（并发布）异常的整个堆栈跟踪，这样可以清楚地识别该行代码。

Answer 2

您必须在aNode设置为null的情况下调用remove。这种行为没有其他解释。

优良作法是断言aNode！= null如果您不期望它。

Answer 3

这可能只意味着aNode本身为空

尝试从链表中删除节点时是否出现NullPointerException？

3 个答案: