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Consider the following class definitions:

public class LinkedChain<T> { private Node<T> firstNode; private int...

X785: Concatenate Linked Chains

Consider the following class definitions:

public class LinkedChain<T> {
    private Node<T> firstNode;
    private int numberOfEntries;

    public LinkedChain() {
        firstNode = null;
        numberOfEntries = 0;
    }// end default constructor


    public Node<T> getFirstNode() {
        return firstNode;
    }


    public int getNumberOfEntries() {
        return numberOfEntries;
    }



    public void push(T newEntry) {
        // TODO Auto-generated method stub
        firstNode = new Node<T>(newEntry, firstNode);
        numberOfEntries++;
    }
}

Where Node is defined as:

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

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


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


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


     public Node getNext() {
         return next;
     }


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

Below, write a Linked Chain method that will take in another LinkedChain object and combine the two chains together. Remember to update the numberOfEntries variable!

For example, if the linked chain looked like this:

A --> B --> C

and the parameter was a linked chain that looked like this:

X --> Y --> Z

this method would change the chain to:

A --> B --> C --> X --> Y --> Z

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