Question

已解决：代码反映解决方案

我一直在处理自定义链接列表，只需使用对列表的一个引用删除具有给定键的节点。

我已经设法用两个引用（节点先前，节点当前）来做到这一点，但是如果仅使用一个引用方法有点困惑。

我的代码适用于除了删除头节点以及尝试删除'88'或不存在'100'的节点时不在列表中的节点（我得到nullpointer异常）的情况。

Here is my test data from the list:
0) 88
1) 2
2) 1
3) 8
4) 11

// Iterative method to delete a node with a given integer key
// Only uses ONE reference variable to traverse the list.
private void delete (int key, Node x) {

    // Check if list is empty
    if (isEmpty()) {
        System.out.println("Cannot delete; the list is empty.");
    }

    // Check if we're deleting the root node
    if (key == head.getKey()) {

        // Now the first in the list is where head was pointing
        removeFromHead();
    }

    // General case: while the next node exists, check its key
    for (x = head; x.getNext() != null; x = x.getNext()) {

        // If the next key is what we are looking for, we need to remove it
        if (key == x.getNext().getKey()) {

            // x skips over the node to be deleted.
            x.putNext(x.getNext().getNext());
        }
    } // End for
}

Answer 1

试试这个：

public Value delete (int key) {

    //check if list is empty
    if (head == null)
        //the key does not exist. return null to let the method caller know
        return null; 

    //check if we're deleting the root node
    if (key == head.getKey()) {
        //set the value of what we're deleting
        Value val = head.getNode().getValue();
        //now the first in the list is where head was pointing
        head = head.getNext();
        //there is now one less item in your list. update the size
        total--;
        //return what we're deleting
        return val;
    }

    // general case: while the next node exists, check its key
    for (Node x = head; x.getNext() != null; x = x.getNext()) {

        //check if the next node's key matches
        if (key == x.getNext().getKey()) {

            //set value of what we're deleting
            Value val = x.getNext().getNode().getValue();

            //x now points to where the node we are deleting points
            x.setNext(x.getNext().getNext());

            //there is now one less item in the list. update the size
            total--;

            //return what we're deleting
            return val; 
        }
    }


    //if we didn't find the key above, it doesn't exist. return null to let the
    // method caller know.
    return null; 
}

这是LinkedList<Value>。一般的想法是存在的，但你必须根据你如何设置一切来定制它。

Answer 2

嗯，你的列表的头部和尾部有问题，因为你没有正确检查它们。您应该在进入while循环之前比较头部的键。循环检查的第一个值是第二个节点（temp.getNext().getKey()），因此您永远不会实际测试头部。此外，在循环结束后，您检查密钥是否在最后一个节点中，您还要调用getNext()。如果确实是最后一个节点，则下一个节点为空，并且没有getKey()方法。

迭代方法：仅使用一个引用变量删除链接列表节点

2 个答案: