在手动链表实现中删除节点中的最后一个对象

Question

我目前对链表的删除操作的实现只有在要删除的节点不在链表末尾时才有效。我需要前一个节点现在指向空，而不是指向最后一个节点，以便将其删除。最后一个节点的删除操作不会抛出 NullPointerException，而是保持列表不变。我将如何更改 else if 条件以反映这一点？

public Book remove(Book bookToRemove){

    Node current = head; //each Node stores a Book

    if(isEmpty()){
        System.out.println("No books to remove!");
    }
    
    //what happens if the next next one is null?
    while(current.getNext() != null && !isEmpty()){
        if(current.getNext().getBook().equals(bookToRemove)){
            Node temp = current.getNext();
            current.setNext(temp.getNext());
            return bookToRemove;
        }

        
        else if(current.getNext() == null){ //
            //if the next node is null, it cannot be removed, or the previous node will have nothing to point to
            current.setNext(null);
            return bookToRemove;
        }

        current = current.getNext();
    }

    return bookToRemove;
}

Answer 1

修改了最初的建议，首先检查头节点，如果是要删除的头节点，则更改为head并退出该功能。流程的其余部分将负责删除列表中的任何其他位置。

public Book remove(Book bookToRemove)
{
    Node current = head; //each Node stores a Book

    if(isEmpty())
    {
        System.out.println("No books to remove!");
        return bookToRemove;   //no need to continue
    }

    if(current.getBook().equals(bookToRemove))
    {
      //found in head itself, make new head and exit
      head = current.getNext();
      return bookToRemove;
    }
    while(current.getNext() != null && !isEmpty())
    {
        if(current.getNext().getBook().equals(bookToRemove))
        {
            Node temp = current.getNext();
            current.setNext(temp.getNext());
            return bookToRemove;
        }        
        current = current.getNext();
    }
    return bookToRemove;
}