Question

In this assignment you are to utilize the Node data structure provided on Blackboard.
In this assignmet you are to write a main program that implements two methods and a main
method as their driver. So, only main and two methods with it. 

Note: You may not use any of the LinkedList class provided on Blackboard, you may use the methods in
      it as a guide (you should actually look at them) but you cannot use any of the methods in your 
      solution. Also, you may not modify the Node class in any way, we will assume that you just used 
      it. You should not submit it. 

You only need to submit only your drive program. 

Implementation Details:
^^^^^^^^^^^^^^^^^^^^^^^
The following are the implementation details of this assignment, you must adhere to all of the 
reauirements. 


Method 1:
^^^^^^^^^
Parameters and return type:
Takes as parameters an array of integers, the size of the array of integer (.length is acceptable also)
and it should return a Node that is supposed to represent the head of a linked list. 

Note: You may also choose to make it a void method and pass the head of the linked list as a parameter, 
      if that's easier for you. You have the choice here. 

Method Logic:
The method is supposed to create a linked list represented by its head and populate it with the 
numbers stored in the array by placing the even numbers in the array first, followed by the odd
numbers. You may not change the order of the numbers inside of the array, the order must remain
the same as it was read from input. 

Example: Assume the array is: [0,1, 4, 6, 7, 9, 2, 10, 11, 14, 13, 19, 20]

         The linked list should be 20->14->10->2->6->4->0->1->7->9->11-13->19

So return back from the function the head of this linked list. 


Method 2:
^^^^^^^^^
Parmeters and return type:
This method shoould take as a parameter the linked list generated in method 1 represented by
its head. 

Method logic:
The method should start by reading in one integer from standard input, and based on that
integer, it needs to shift the linked list by that many positions. Keep in mind that you need
to do the necessary error checking, the shifting can be between 0 and the size of the linked list - 1. 

Example:
Assume the given linked list is:  20->14->10->2->6->4->0->1->7->9->11-13->19

You read in an integer: You input the number 3. 
The linked list should look like: 2->6->4->0->1->7->9->11-13->19->20->14->10

If you read in a 6:
The linked list should look like: 0->1->7->9->11-13->19->20->14->10->2->6->4

If you read in a 13 The method should print an error asking you to enter a number between 0-12.


The main program:
^^^^^^^^^^^^^^^^^
1. Your program should run and ask the user to input the size of an array of integers. Once that's
   done, the program should read these integers and store them into an array. 

2. Once the array has been populated and its size is known, then you need to call method 1 defined 
   above. The result should be a head pointer of a linked list. 
   
   At this point you declare a cursor method and go through the linked list and print it to the screen.
   Based on the above example: 20 14 10 2 6 4 0 1 7 9 11 13 19.

3. Call method 2

4. Print the linked list resulting from calling method 2. The rotated linked list.

Here's the shell of my code

public static void main(String [] args)
{
    Node<Integer> head;

    head = buildList();
    shiftList(head);
}

public static Node<Integer> buldList()
{
    // declare your scanner in
    // declare the number of integers you're going to read (n)
    // declare your array a

    Node<Integer> head, cursor;

    for(i = 0; i < n; i++ ){

        a[i] = in.nextInt();

        // if i == 0 that means that the head and the cursor are both
        // pointing to null and there exists no nodes in the linked list
        // therefore we need to create our first node first, and have the
        // head and the cursor both point to it.
        // After that i will not be equal to zero and we can then start executing
        // the else part of this if statement by checking if the number is odd
        // or even.
        if( i == 0){
            head = new Node<Integer>(a[i], null);
            // insert the code that makes the cursor point to the head

        } else {
            if(a[i] % 2 == 1){   // if the number is odd

                cursor.setNext(new Node(a[i], null));
                // Insert the code that moves the cursor over.

            }else{
                // the number is even, so we will insert at the head and move the head
                // to point to the newly inserted node.
                // Use a code that is highly similar to the listHeadInsert
                // method in the linked list class. Do not call the listHeadInsert
                // method though, but you can use its logic

            }
        }
    }

    return head;
}


public static void shiftList(Node<Integer> head)
{
    // in this code you will need to have some temporary variables starting

    // Ask the user to enter a number by which to shift
    // make the necessary error checking to make sure that the shift is valid
    // call this number x.

    tmp = head;
    cursor = head;

    // now you moved tmp to the location where head needs to be
    for(i = 0; i < x - 1; i++)
        tmp = tmp.getNext();

    // at this point find the end of the list
    while(cursor.getNext() != null)
        cursor = cursor.getNext();

    // make the last node point to the old head
    cursor.setNext(head);

    // Now write the code that moves the head over and set the Next of the node
    // before the new location of the head to null.



}

public class Node<T>
{
   
    private T value;            // this is the data value
    private Node<T> next; // this is pointing to the next node


    // the node constructor
    public Node (T v, Node<T> n)
        {
            value = v;
            next = n;
        }

    // getters and setters for the node's value and next pointer
    public T getValue() {return value;}
    public Node<T> getNext() {return next;}
    public void setValue(T v){value = v;}
    public void setNext(Node<T> n){next = n;}

}

Answer 1

import java.util.Scanner;

public class LinkedListDemo {

    public static void main(String [] args)
    {
        Node<Integer> head;

        head = buildList();
        printList(head);

        System.out.println();

        head = shiftList(head);
        printList(head);
    }

    public static void printList(Node<Integer> node) {
        while(node != null) {
            System.out.print(node.getValue() + " ");
            node = node.getNext();
        }
        System.out.println(); 

    }

    public static Node<Integer> buildList()
    {
        // declare your scanner in
        // declare the number of integers you're going to read (n)
        // declare your array a
        Scanner in = new Scanner(System.in);
        int n = in.nextInt();
        int a[] = new int[n];

        int i;

        Node<Integer> head = null, cursor = null;

        for(i = 0; i < n; i++ ){

            a[i] = in.nextInt();

            // if i == 0 that means that the head and the cursor are both
            // pointing to null and there exists no nodes in the linked list
            // therefore we need to create our first node first, and have the
            // head and the cursor both point to it.
            // After that i will not be equal to zero and we can then start executing
            // the else part of this if statement by checking if the number is odd
            // or even.
            if( i == 0){
                head = new Node<Integer>(a[i], null);
                // insert the code that makes the cursor point to the head
                cursor = head;
            } else {
                if(a[i] % 2 == 1){   // if the number is odd

                    cursor.setNext(new Node<Integer>(a[i], null));
                    // Insert the code that moves the cursor over.
                    cursor = cursor.getNext();
                }else{
                    // the number is even, so we will insert at the head and move the head
                    // to point to the newly inserted node.
                    // Use a code that is highly similar to the listHeadInsert
                    // method in the linked list class. Do not call the listHeadInsert
                    // method though, but you can use its logic
                    head = new Node<Integer>(a[i], head);
                }
            }
        }

        return head;
    }


    public static Node<Integer> shiftList(Node<Integer> head)
    {
        // in this code you will need to have some temporary variables starting

        // Ask the user to enter a number by which to shift
        // make the necessary error checking to make sure that the shift is valid
        // call this number x.
        Scanner in = new Scanner(System.in);
        int x = in.nextInt();
        int i;

        Node<Integer> tmp = head;
        Node<Integer> cursor = head;

        // now you moved tmp to the location where head needs to be
        for(i = 0; i < x - 1; i++) {
            tmp = tmp.getNext();
            if(tmp == null) {
                System.out.println("Invalid Shift entered.");
                return head;
            }
        }

        // at this point find the end of the list
        while(cursor.getNext() != null)
            cursor = cursor.getNext();

        // make the last node point to the old head
        cursor.setNext(head);

        // Now write the code that moves the head over and set the Next of the node
        // before the new location of the head to null.
        head = tmp.getNext();
        tmp.setNext(null);

        return head;
    }

}

class Node<T>
{

    private T value;            // this is the data value
    private Node<T> next; // this is pointing to the next node


    // the node constructor
    public Node (T v, Node<T> n)
        {
            value = v;
            next = n;
        }

    // getters and setters for the node's value and next pointer
    public T getValue() {return value;}
    public Node<T> getNext() {return next;}
    public void setValue(T v){value = v;}
    public void setNext(Node<T> n){next = n;}

}

Please upvote, as i have given the exact answer as asked in question. Still in case of any concerns in code, let me know in comments. Thanks!

▲ ■ <terminated> LinkedListDemo [Java licatic System.out.println("Invalid Shift entered."); return head; 10 1 2 3 4 56789 10 10 8 6 421 3 57 9 97 98 100 101 102 103 104 105 106 107 108 109 110 // at this point find the end of the list while(cursor.getNext) !-null) tmp: 6 cursor: 9 4 2 1 3 579 10 8 6 cursor-cursor.getNext); // make the last node point to the old head cursor.setNext (head); // Now write the code that moves the head over and set the Next of the node // before the new location of the head to null. head-tmp.getNext); tmp.setNext (nul1); 112 113 114 115 116 return head; 118 class Node<T> 119 ( 120 121 122 private Node<T> next; // this is pointing to the next node 123 124 125 // t 126 public Node (T v, Node<T> n) 127 128 129 130 131 132 // getters and setters for the node's value and next pointer 133 public T getValue() (return value;) 134 public Node<T> getNext) (return next; 135 public void setValue(T v)value- v;) 136 public void setNext (Node<T> n)next n; 137 private T value; // this is the data value he node constructor value = v; next n;