CODEWITHSUNDEEP
javarecursionnodessingly-linked-list
Adan Vivero
Adan Vivero

Reputation: 422

How to create a remove method using recursion with an index in the parameter in Java?

I'm having trouble on how to start this method. I am trying to create a remove method using recursions in my code. Basically I have a public and private remove method. The remove(int) method, which is public, should remove the element at the specified index in the list. I need to address the case in which the list is empty and/or the removed element is the first in the list. If the index parameter is invalid, an IndexOutOfBoundsException should be thrown. To allow for a recursive implementation, this method should address special cases and delegate to remove(int, int, Node) for recursion.

Here's the class:

public class SortedLinkedList<E extends Comparable<E>>
{
    private Node first;
    private int size;
    // ...
}

And here's the code: 

public void remove(int index)
{
    if(index < 0 || index > size)
    {
        throw new IndexOutOfBoundsException();
    }
    remove(index++, 0, first);
    if (index == 0)
    {
        if(size == 1)
        {
            first = null;
        }
        else
        {
            first = first.next;
        }
    }
    size--;
}

And the private method:

private void remove(int index, int currentIndex, Node n)
{
    if(index == currentIndex)
    {
        remove(index, currentIndex, n.next);
    }
    remove(index, currentIndex, n.next.next);
}

With a private class: 

private class Node
{
    private E data;
    private Node next;

    public Node(E data, Node next)
    {
        this.data = data;
        this.next = next;
    }
}

Upvotes: 0

Views: 982

Answers (1)

Kaue Silveira
Kaue Silveira

Reputation: 170

Returning void

Using two indexes

private void remove(int index, int current, Node n) {
  if (n == null || index <= 0 || (index == 1 && n.next == null) {
    throw new IndexOutOfBoundsException();
  }
  if (current == index - 1) {
    // Remove 'n.next'.
    n.next = n.next.next; 
  } else {
    remove(index, current + 1, n.next);
  }
}

Usage

public void remove(int index) {
  if (first == null || index < 0) {
    throw new IndexOutOfBoundsException();
  }
  if (index == 0) {
    // Remove 'first'.
    first = first.next;
  } else {
    remove(index, 0, first);
  }
  size--;
}

Using one index

Only one index is needed:

private void remove(int index, Node n) {
  if (n == null || index <= 0 || (index == 1 && n.next == null) {
    throw new IndexOutOfBoundsException();
  }
  if (index == 1) {
    // Remove 'n.next'.
    n.next = n.next.next; 
  } else {
    remove(index - 1, n.next);
  }
}

Usage

public void remove(int index) {
  if (first == null || index < 0) {
    throw new IndexOutOfBoundsException();
  }
  if (index == 0) {
    // Remove 'first'.
    first = first.next;
  } else {
    remove(index, first);
  }
  size--;
}

Returning Node

Even better is to return Node instead of void:

private Node remove(int index, Node n) {
  if (n == null || index < 0) {
    throw new IndexOutOfBoundsException();
  }
  if (index == 0) {
    // Remove 'n' and return the rest of the list.
    return n.next; 
  }
  // 'n' stays. Update the rest of the list and return it.
  n.next = remove(index - 1, n.next);
  return n;
}

Usage

public void remove(int index) {
  first = remove(index, first);
  size--;
}

Upvotes: 1

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