JAVA Queue Implementation using Arrays

Question

I'm trying to implement Queues in JAVA. I'm a beginner. I dont understand why this isn't working. Push() works fine but pop() isn't working. Can someone please point out where im going wrong?

pop():

 public void pop()
    {
    for(int i=0;i<length;i++)
    {
         while(i<(length-1))
         {
        arr[i]=arr[i+1];

        }

    }

    }

push():

public void push(int x)
    {
    push:for(int i=0;i<length;i++)
    {
        if(arr[i]==null) 
        {
        arr[i]=x;
        break push;
        }
    }
    }

show():

public void show()
    {
    int c=0;
    for(int i=0;i<length;i++)
        //if(arr[i]!=null) 
        {
            System.out.println(arr[i]);
            c++;

        }
    System.out.println("Current Capacity "+c+"/"+length);       
    }

main()

public static void main(String...i)
    {
        System.out.println("Stack Implementation");
        Queue stack = new Queue();
        System.out.println("Push");
        stack.push(1);
        stack.push(2);
        stack.push(3);
        stack.push(4);
        stack.push(5);
        stack.show();
        System.out.println("Pop");
        stack.pop();
        stack.show();
    }

The output doesn't show any data after pop() is run.

Please post the whole code, e.g. the definition of the variables i, arr and so on. — mrbela
– mrbela, Commented Dec 18, 2018 at 16:17
Just a clarification on terminology. Queues are typically queue() and dequeue(), stacks use push() and pop(). — Jason Armstrong
– Jason Armstrong, Commented Dec 18, 2018 at 16:19
Please explain "isn't working". What did you expect to happen, why did you expect it, and what happened instead (error/exception/other result)? — Pshemo
– Pshemo, Commented Dec 18, 2018 at 16:21
Updated the whole code - this is my first post on stack overflow - sorry for confusing with partial code. — Chandradhar
– Chandradhar, Commented Dec 18, 2018 at 16:23

Thomas · Accepted Answer · 2018-12-18 16:24:44Z

1

You don't increment i in pop() so the while loop will run endlessly.

In push you are using a for loop which increments i: :for(int i=0;i<length;i++ /*here*/)

You also don't initialize i in pop() so it will probably have the value of the last increment in push(). That value will be the index of the next empty element (if there's one left) and thus that's wrong anyways. However, you want to pop from the front, so you'd need to start at i = 0 - in that case another for loop would work as well, i.e. you just copy the value of element at i+1 to the index i and set the last element to null (for more efficiency you could stop once i+1 has a null element).

Edit: now that you've posted more code for pop() the situation is a little different. You are already using a for loop in pop() but another loop inside that. I assume you want to do if(i<(length-1)) instead of while(i<(length-1)) - but in that case you'd still have to handle the last element, i.e. once the queue was full you'd need to set the last element to null when you pop one and move the rest.

edited Dec 18, 2018 at 16:24

answered Dec 18, 2018 at 16:19

Thomas

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2 Comments

Chandradhar Over a year ago

my logic: a[0] will become a[1], a[1] will become a[2] and before a[10] becomes a[11] (outofbounds), the loop stops(length-1). Isn't this correct?

Thomas Over a year ago

@Chandradhar your logic is sound but your implementation is faulty. If you change while(...) with if(...) you'll get the logic you want (while is another loop keyword, i.e. the loop keeps running until the condition is met, which will never happen). The only problem with your logic would be that if you stop before a[10] = a[11] in your example a[10] would never change - i.e. you need a a[10] = null after the loop (this could be more efficient but it's a start).

Davide Lorenzo MARINO · Accepted Answer · 2018-12-18 16:19:33Z

0

When you pushed the element you need to return from the method:

 public void push(int x) {
   for (int i = 0; i < length; i++) {
     if (arr[i] == null) {
       arr[i] = x;
       return;  // Exit from push when you added the element in the right position
     }
   }
 }

Note that this code is not optimized. Push an element requires O(n), so can waste a lot of time for big queues, but this is the closest solution to your code. Anyway a simple optimization can be done introducing a variable holding the last used index. So you can use that variable to push and pop an element in O(1).

answered Dec 18, 2018 at 16:19

Davide Lorenzo MARINO

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1 Comment

Chandradhar Over a year ago

my code for push is working fine with a break statement in place of return. The code for pop() isn't working. But I'd love to know if the return statement is better than break? or both okay to use in this scenario?

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