1

I wrote the following implementation of the queue using a linked list that does not maintain a reference to the tail node. When I try to print the queue, it outputs only the head i.e. only one node. What is the error? Thanks in advance!

package DataStructures;

import java.util.Scanner;

class Node {
    int x;
    Node nextNode;

    public Node(int x) {
        this.x = x;
        nextNode = null;
    }
}

class Queue {
    Node head = null;
    int n = 0;

    public void enqueue(int x) {
        if (n==0){
            head = new Node(x);
            n++;
            return;
        }
        Node tempHead = head;
        while (tempHead != null){
            tempHead = tempHead.nextNode;
        }
        tempHead = new Node(x);
        tempHead.nextNode = null;
        n++;
    }

    public int dequeue() {
        if (head == null) {
            throw new Error("Queue under flow Error!");
        } else {
            int x = head.x;
            head = head.nextNode;
            return x;
        }
    }

    public void printTheQueue() {
        Node tempNode = head;
        System.out.println("hi");
        while (tempNode != null){
            System.out.print(tempNode.x + "  ");
            tempNode = tempNode.nextNode;
        }

    }

}

public class QueueTest {

    private static Scanner in = new Scanner(System.in);

    public static void main(String[] args) {
        Queue queue = new Queue();
        while (true){
            int x = in.nextInt();
            if (x == -1){
                break;
            } else{
                queue.enqueue(x);
            }
        }

        queue.printTheQueue();
    }

}
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2
  • When you enqueue, there is nothing that connects between the head and your new node. Commented Aug 15, 2015 at 10:59
  • @RealSkeptic But I do have a temporary reference to head and propagate through it to reach the last node. Once I reach the last node, I point its nextNode to new node with data key x. Commented Aug 15, 2015 at 11:02

2 Answers 2

Reset to default
2

You never assign a node to nextNode, so your list is either empty or consists of one node.

Here is a solution:

public void enqueue(int x) {
    n++;
    if (head == null) {
        head = new Node(x);
    else {
        Node last = head;
        while (last.nextNode != null)
            last = last.nextNode;
        last.nextNode = new Node(x);
    }
}

Technically you don't need n but you could use it as cache for the size of the list. And you should decrease it in deque().

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3 Comments

I would be thankful if you tell me what should be the change in the code
Thanx... got it... I was assigning new Node() to tempHead which was a flying reference.. instead I should have assigned new Node() to tempHead.nextNode with proper logic!!!
The n++ should come at the end of the method.
0

Make your enqueue to this:

public void enqueue(int x) {
 if (n==0){
        head = new Node(x);
        head.nextNode=null;
        n++;
        return;
    }
    Node tempHead = head;
    while (tempHead.nextNode!= null){
        tempHead = tempHead.nextNode;
    }
    Node newNode = new Node(x);
    tempHead.nextNode=newNode;
    newNode.nextNode = null;
    n++;
}
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1 Comment

Yes its correct!......but if I remove the if statement if(n==0), your code would crash since head is null and we are accessing its nextNode.... Can You provide a hint for this??

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