Can someone please explain how can I remove first and last nodes in my doubly linked list?

Question

I have created a class SLList where I take in a doubly linked list SLList and make some adaptations to it.

Now, I'm trying to remove the first node in the doubly linked list and the last node in the doubly linked list via two functions known as public T firstout() and public T lastout() and I think the purpose of each function is pretty self-explanatory.

Based on the recommendation of StackOverflow user, I have decided to use just one sentinel to aid in my task. However, I'm struggling to rightfully implement sentinels to my advantage.

Here is my code so far:

public class SLList<T>{ // Add Generics here
    private class IntNode {
        private T data;
        private IntNode previous;
        private IntNode next;

        public IntNode (T data, IntNode previous, IntNode next) {
            this.data = data;
            this.previous = previous;
            this.next = next;
        }

        public IntNode () { // Self-referencing node
            next = previous = this;
        }
    }

    IntNode sentinel; // One sentinel can serve both purposes
    private int length = 0;

    public SLList(){
        sentinel = new IntNode(); // Self referencing previous/next
    }

    public void addFirst(T data) {
        IntNode node = new IntNode(data, sentinel, sentinel.next);
        sentinel.next = node;
        node.next.previous = node;
    }
    public boolean isempty() {
        return length == 0;
    }


    public T firstout() {
    return sentinel.previous == null;

    }
    public T lastout() {
    return sentinel.next == null;
    }

Based on what I've tried to find, I thought .previous points to the start of the list and setting it to null would effectively remove the first one and .next points to the next node, and setting it to null would also remove the last one.

Where has my understanding gone wrong here? Any tips or code improvements would be greatly appreciated

score 1 · Accepted Answer · answered Oct 04 '22 at 17:56

First of all, you don't need length to know whether the list is empty. You could also define the isEmpty method like this:

    public boolean isEmpty() {
        return sentinel.next == sentinel;
    }

NB: I prefer camelCase instead of all lowercased method names, so isEmpty instead of isempty, and firstOut instead of firstout.

For the firstOut and lastOut methods, I would first define a private method that removes a given node from the list (assuming it is in the list):

    private T removeNode(IntNode node) {
        // Link neighbors to eachother, skipping over given node
        node.previous.next = node.next;
        node.next.previous = node.previous;
        return node.data;
    }

This can then be used as follows:

    public T firstOut() {
        return removeNode(sentinel.next);
    }
    
    public T lastOut() {
        return removeNode(sentinel.previous);
    }

In this version, if the list is empty and firstOut or lastOut is called, the removeNode call will not change anything to the list, because the node it gets will be the sentinel node which links to itself, and so both assignments just confirm what is already the case. firstOut or lastOut will return the value of the sentinel node in that case.

You could alternatively throw an exception when the list is empty when removeNode is called. But exception handling is a topic on itself, so I will not dive into that here.

Can someone please explain how can I remove first and last nodes in my doubly linked list?

1 Answers1