在链接列表中删除节点

Question

尝试按索引删除节点。我想用索引打印出节点列表，以便用户可以选择索引。我认为我的逻辑可以使用索引打印列表但没有输入出来:(

在愚弄这一点时，我仍然无法打印节点列表，但是“您要删除的索引中的索引中的索引”是输出的，并且能够占用用户的选择，但最终得到了nullpointerexception。

            else if (menu.equals("d")) {
            EntryNode temp = head;
            while (temp != null) {
                for (int i = 0; i < addressBook.length(); i++) {
                    //gets node at index
                    System.out.println(temp.getFirstName() + i);
                    temp = temp.getNext();
                }
            System.out.println(" ");
            System.out.println("Please enter the index of the entry you wish to delete ");
            int index = keyboard.nextInt();
            addressBook.removeEntry(index);
            }

        }

.

删除方法： 公共void removeEntry（int索引）{

    //delete from the head
    if (index == 0) {
        EntryNode temp = head;
        head = temp.getNext();
        temp.setNext(null);
        head.setPrev(null);
        size--;
    }
    //delete from the tail
    else if (index == length()) {
        EntryNode temp = tail;
        temp.setPrev(null);
        tail.setNext(null);
        tail = temp.getPrev();

        size--;
    }
    //in the middle
    else {
        EntryNode temp = head;
        for (int i = 0; i < index; i++) {
            //gets node at index
            temp = temp.getNext();
        }
        //set node after temp's previous to temp's previous 
        temp.getNext().setPrev(temp.getPrev());
        temp.getPrev().setNext(temp.getNext());
        temp.setNext(null);
        temp.setPrev(null);
        size--;
    }
}

.

nullpointerexception来自：

//set node after temp's previous to temp's previous 
temp.getNext().setPrev(temp.getPrev());

.

Answer 1

You should check if temp.getNext() is not null before calling the setPrev() on it.

Also, you should check with length()-1 as you have nodes which are zero indexed.

Answer 2

Indices are in the range [0, length() - 1], so you should use

else if (index == length() - 1) {

instead.