Java使用带有LinkedList的节点

Question

I've been working through some standard coding interview questions from a book I recently bought, and I came across the following question and answer:

Implement an algorithm to find the nth to last element in a linked list.

Here's the provided answer:

public static LinkedListNode findNtoLast(LinkedListNode head, int n) { //changing LinkedListNode to ListNode<String>
        if(head == null || n < 1) {
            return null;
        }
        LinkedListNode p1 = head;
        LinkedListNode p2 = head;
        for(int j = 0; j < n-1; ++j) {
            if(p2 == null) {
                return null;
            }
            p2 = p2.next;
        }
        if(p2 == null) {
            return null;
        }
        while(p2.next != null) {
            p1 = p1.next;
            p2 = p2.next;
        }
        return p1;
    }

I understand the algorithm, how it works, and why the book lists this as its answer, but I'm confused about how to access the LinkedListNodes to send as an argument to the method. I know that I'd have to create a LinkedListNode class (since Java doesn't already have one), but I can't seem to figure out how to do that. It's frustrating because I feel like I should know how to do this. Here's something that I've been working on. I'd greatly appreciate any clarification. You can expand/comment on my code or offer your own alternatives. Thanks.

class ListNode<E> {
    ListNode<E> next;
    E data;

    public ListNode(E value) {
        data = value;
        next = null;
    }

    public ListNode(E value, ListNode<E> n) {
        data = value;
        next = n;
    }

    public void setNext(ListNode<E> n) {
        next = n;
    }
}

public class MyLinkedList<E> extends LinkedList {
    LinkedList<ListNode<E>> list;
    ListNode<E> head;
    ListNode<E> tail;
    ListNode<E> current;
    ListNode<E> prev;

    public MyLinkedList() {
        list = null;
        head = null;
        tail = null;
        current = null;
        prev = null;
    }

    public MyLinkedList(LinkedList<E> paramList) {
        list = (LinkedList<ListNode<E>>) paramList; //or maybe create a loop assigning each ListNode a value and next ptr
        head = list.getFirst();
        tail = list.getLast(); //will need to update tail every time add new node
        current = null;
        prev = null;
    }

    public void addNode(E value) {
        super.add(value);
        //ListNode<E> temp = tail;
        current = new ListNode<E>(value);
        tail.setNext(current);
        tail = current;
    }

    public LinkedList<ListNode<E>> getList() {
        return list;
    }

    public ListNode<E> getHead() {
        return head;
    }

    public ListNode<E> getTail() {
        return tail;
    }

    public ListNode<E> getCurrent() {
        return current;
    }

    public ListNode<E> getPrev() {
        return prev;
    }


}

How can the LinkedListNode head from a LinkedList?

Update: I think part of my confusion comes from what to put in the main method. Do I need to create a LinkedList of ListNode? If I do that, how would I connect the ListNodes to each other? How would I connect them without using a LinkedList collection object? If someone could show me how they would code the main method, I think that would put things into enough perspective for me to solve my issues. Here's my latest attempt at the main method:

public static void main(String args[]) {
        LinkedList<ListNode<String>> list = new LinkedList<ListNode<String>>();
        //MyLinkedList<ListNode<String>> list = new MyLinkedList(linkedList);
        list.add(new ListNode<String>("Jeff"));
        list.add(new ListNode<String>("Brian"));
        list.add(new ListNode<String>("Negin"));
        list.add(new ListNode<String>("Alex"));
        list.add(new ListNode<String>("Alaina"));
        int n = 3;
        //ListIterator<String> itr1 = list.listIterator();
        //ListIterator<String> itr2 = list.listIterator();
        LinkedListNode<String> head = new LinkedListNode(list.getFirst(), null);
        //String result = findNtoLast(itr1, itr2, n);
        //System.out.println("The " + n + "th to the last value: " + result);
        //LinkedListNode<String> nth = findNtoLast(list.getFirst(), n);
        ListNode<String> nth = findNtoLast(list.getFirst(), n);
        System.out.println("The " + n + "th to the last value: " + nth);
    }

In an attempt to connect the nodes without using a custom linked list class, I have edited my ListNode class to the following:

class ListNode<E> {
    ListNode<E> next;
    ListNode<E> prev; //only used for linking nodes in singly linked list
    ListNode<E> current; //also only used for linking nodes in singly linked list
    E data;
    private static int size = 0;

    public ListNode() {
        data = null;
        next = null;
        current = null;
        if(size > 0) { //changed from prev != null because no code to make prev not null
            prev.setNext(this);
        }
        size++;
    }
    public ListNode(E value) {
        data = value;
        next = null;
        current = this;
        System.out.println("current is " + current);
        if(size > 0) {
            prev.setNext(current);//this line causing npe
        }
        else
        {
            prev = current;
            System.out.println("prev now set to " + prev);
        }
        size++;
        System.out.println("after constructor, size is " + size);
    }

    public ListNode(E value, ListNode<E> n) {
        data = value;
        next = n;
        current = this;
        if(size > 0) {
            prev.setNext(this);
        }
        size++;
    }

    public void setNext(ListNode<E> n) {
        next = n;
    }
}

As is right now, the program will run until it reaches prev.setNext(current); in the single argument constructor for ListNode. Neither current nor prev are null at the time this line is reached. Any advice would be greatly appreciated. Thanks.

Answer 1

You don't actually need a separate LinkedList class; the ListNode class is a linked list. Or, to state it differently, a reference to the head of the list is a reference to the list.

The use of head, tail, current, prev in the sample code you posted has come from a double-linked list which is a data type that has links in both directions. This is more efficient for certain types of applications (such as finding the nth last item).

So I would recommend renaming your ListNode class to LinkedList and renaming next to tail .

To add a new item to the list you need a method that creates a new list with the new item at it's head. Here is an example:

class LinkedList<E> {
    ...

    private LinkedList(E value, LinkedList<E> tail) {
        this.data = value;
        this.tail = tail;
    }

    public LinkedList<E> prependItem(E item) {
        return new LinkedList(item, this);
    }
}

Then to add a new item i to list you use list = list.prependItem(i);

If for some reason you need to always add the items to the end, then:

private LinkedList(E value) {
    this.data = value;
    this.tail = null;
}

public void appendItem(E item) {
    LinkedList<E> list = this;
    while (list.tail != null)
        list = list.tail;
    list.tail = new LinkedList<>(item);
}

However this is obviously pretty inefficient for long lists. If you need to do this then either use a different data structure or just reverse the list when you have finished adding to it.

Incidentally, an interesting side effect of this is that a reference to any item in the list is a reference to a linked list. This makes recursion very easy. For example, here's a recursive solution for finding the length of a list:

public int getLength(LinkedList list) {
    if (list == null) {
        return 0;
    } else {
        return 1 + getLength(list.getTail());
    }
}

And using this a simple (but very inefficient!) solution to the problem you provided - I've renamed the method to make its function more obvious:

public LinkedList getTailOfListOfLengthN(LinkedList list, int n) {
    int length = getLength(list);
    if (length < n) {
        return null;
    } else if (length == n) {
        return list;
    } else {
        return getTailOfLengthN(list.getTail(), n);
    }
}

And to reverse the list:

public LinkedList<E> reverse() {
    if (tail == null) {
        return this;
    } else {
        LinkedList<E> list = reverse(tail);
        tail.tail = this;
        tail = null;
        return list;
    }
}

As I hope you can see this makes the methods a lot more elegant than separating the node list classes.

Answer 2

Actually you have created a linked list with you class ListNode.

A linked list is made of a node and a reference to another linked list (see the recursion?).