Question

(Please use Java Eclipse)

QUESTION 1:

For the question below, assume the following implementation of LinkedQueue:

public static final class LinkedQueue<T> implements QueueInterface<T> {
    private Node<T> firstNode;
    private Node<T> lastNode;

    public LinkedQueue() {
        firstNode = null;
        lastNode = null;
    }

    @Override
    public T getFront() {
        if (isEmpty()) {
            return null;
        }
        return firstNode.getData();
    }

    @Override
    public boolean isEmpty() {
        return firstNode == null;
    }

    @Override
    public void clear() {
        firstNode = null;
        lastNode = null;

    }

    public class Node<E> {
        private E data; // Entry in bag
        private Node<E> next; // Link to next node

        public Node(E dataPortion) {
            this(dataPortion, null);
        } // end constructor

        public Node(E dataPortion, Node<E> nextNode) {
            data = dataPortion;
            next = nextNode;
        } // end constructor

        public E getData() {
            if (data != null) {
                return data;
            }
            return null;
        }

        public Node<E> getNext() {
            return next;
        }

        public void setNext(Node<E> newNext) {
            next = newNext;
        }

} // end LinkedQueue

Below, finish creating the enqueue method that has been started for you.

public void enqueuePractice(T newEntry) {

    Node<T> newNode = new Node<T>(newEntry);

    if (isEmpty()) {

        firstNode = newNode;

    }

}

QUESTION 2:

For the question below, assume the following implementation of LinkedQueue:

public static final class LinkedQueue<T> implements QueueInterface<T> {
    private Node<T> firstNode;
    private Node<T> lastNode;

    public LinkedQueue() {
        firstNode = null;
        lastNode = null;
    }

    @Override
    public T getFront() {
        if (isEmpty()) {
            return null;
        }
        return firstNode.getData();
    }

    @Override
    public boolean isEmpty() {
        return firstNode == null;
    }

    @Override
    public void clear() {
        firstNode = null;
        lastNode = null;

    }

    public class Node<E> {
        private E data; // Entry in bag
        private Node<E> next; // Link to next node

        public Node(E dataPortion) {
            this(dataPortion, null);
        } // end constructor

        public Node(E dataPortion, Node<E> nextNode) {
            data = dataPortion;
            next = nextNode;
        } // end constructor

        public E getData() {
            if (data != null) {
                return data;
            }
            return null;
        }

        public Node<E> getNext() {
            return next;
        }

        public void setNext(Node<E> newNext) {
            next = newNext;
        }

} // end LinkedQueue

Your task is to create a "priority" enqueue method. If something is enqueued, it normally goes to the back of the line, but in this method it will go to the front of the line.

For example, if you had a linked queue that looked like:

(firstNode) 1 --> 2 --> 3 (lastNode)

and you ran priorityEnqueue(4)

the result would be:

(firstNode) 4 --> 1 --> 2 --> 3 (lastNode)

public void priorityEnqueue(T newEntry) {

    Node<T> newNode = new Node<T>(newEntry);

    if (isEmpty()) {

        lastNode = newNode;

    } 

}

QUESTION 3:

For the question below, assume the following implementation of LinkedQueue:

public static final class LinkedQueue<T> implements QueueInterface<T> {
    private Node<T> firstNode;
    private Node<T> lastNode;

    public LinkedQueue() {
        firstNode = null;
        lastNode = null;
    }

    @Override
    public T getFront() {
        if (isEmpty()) {
            return null;
        }
        return firstNode.getData();
    }

    @Override
    public boolean isEmpty() {
        return firstNode == null;
    }

    @Override
    public void clear() {
        firstNode = null;
        lastNode = null;

    }

    public class Node<E> {
        private E data; // Entry in bag
        private Node<E> next; // Link to next node

        public Node(E dataPortion) {
            this(dataPortion, null);
        } // end constructor

        public Node(E dataPortion, Node<E> nextNode) {
            data = dataPortion;
            next = nextNode;
        } // end constructor

        public E getData() {
            if (data != null) {
                return data;
            }
            return null;
        }

        public Node<E> getNext() {
            return next;
        }

        public void setNext(Node<E> newNext) {
            next = newNext;
        }

} // end LinkedQueue

Your task is to create an "almostPriority" enqueue method. If something is enqueued, it normally goes to the back of the queue, but in this method it will be set immediately after the first element in the queue.

For example, if you had a linked queue that looked like:

(firstNode) 1 --> 2 --> 3 (lastNode)

and you ran almostPriorityEnqueue(4)

the result would be:

(firstNode) 1 --> 4 --> 2 --> 3 (lastNode)

@SuppressWarnings("unchecked")

public void almostPriorityEnqueue(T newEntry) {

    Node<T> newNode = new Node<T>(newEntry);

    if (isEmpty()) {

        lastNode = newNode;

        firstNode = newNode;

    }

}

​

Answer 1

Here is the completed code for each of the required methods for questions 1, 2 and 3. Comments are included, go through it, learn how things work and let me know if you have any doubts or if you need anything to change. If you are satisfied with the solution, please rate the answer. If not, PLEASE let me know before you rate, I’ll help you fix whatever issues. Thanks

//enqueue method for question 1
public void enqueuePractice(T newEntry) {
        // creating a new node
        Node<T> newNode = new Node<T>(newEntry);
        // if queue is empty, adding as both first and last node
        if (isEmpty()) {
                firstNode = newNode;
                lastNode = newNode;
        } else {
                // otherwise appending next of last node and setting as new last
                // node.
                lastNode.setNext(newNode);
                lastNode = newNode;
        }

}

// priorityEnqueue method for question 2
public void priorityEnqueue(T newEntry) {
        Node<T> newNode = new Node<T>(newEntry);
        // if queue is empty, adding as both first and last node
        if (isEmpty()) {
                firstNode = lastNode = newNode;
        } else {
                // otherwise adding before current firstNode and setting as new
                // firstNode
                newNode.setNext(firstNode);
                firstNode = newNode;
        }
}

// almostPriorityEnqueue method for question 3
@SuppressWarnings("unchecked")
public void almostPriorityEnqueue(T newEntry) {
        Node<T> newNode = new Node<T>(newEntry);
        // if queue is empty, adding as both first and last node
        if (isEmpty()) {
                lastNode = newNode;
                firstNode = newNode;
        } else {
                // adding between firstNode and firstNode.next
                newNode.setNext(firstNode.getNext());
                firstNode.setNext(newNode);
                // if next is null, setting newNode as last node.
                if (newNode.getNext() == null) {
                        lastNode = newNode;
                }
        }

}