> ### Solution for
> Programming Exercise 11.3

> THIS PAGE DISCUSSES ONE POSSIBLE SOLUTION to the following exercise from this on-line Java textbook.

> Exercise 11.3: Suppose that linked lists of integers are made from objects belonging to the class

>              class ListNode {
>                 int item;       // An item in the list.
>                 ListNode next;  // Pointer to the next node in the list.
>              }

> Write a subroutine that will make a copy of a list, with the order of the items of the list reversed. The subroutine should have a parameter of type `ListNode`, and it should return a value of type `ListNode`. The original list should not be modified.

> You should also write a `main()` routine to test your subroutine.

> Discussion

> To make any linked list from scratch, you have to create nodes one-by-one and link them together. In this case, we want to make nodes that contain copies of the items from the original list. We can run through the original list, look at each item, create a new node that contains a copy of that item, and link that new node into the reversed list that we are constructing. We just have to make sure that the nodes in the new list are in the desired order.

> In fact this is pretty easy: As we run down the original list from start to finish, we need to build the reversed list from back to front. The first item in the original list should be at the back of the reversed list, the second item from the original goes in front of that item, and so on. This is equivalent to "pushing" the items onto the reversed list, using the same push operation that is used for a stack. An algorithm for this is:

>               Let rev be an empty list
>               for each item in the original list:
>                   Push the item onto rev
>                   Move on to the next item

> This can be coded into the subroutine we need as follows:

>           static ListNode reverse( ListNode list ) {
>                   // Return a new list containing the same items as the list,
>                   // but in the reverse order.
>              ListNode rev = null;     // rev will be the reversed list.
>              ListNode runner = list;  // For running through the nodes of list.
>              while (runner != null) {
>                     // "Push" the next node of list onto the front of rev.
>                 ListNode newNode = new ListNode();
>                 newNode.item = runner.item;
>                 newNode.next = rev;
>                 rev = newNode;
>                    // Move on to the next node in the list.
>                 runner = runner.next;
>              }
>              return rev;
>           } // end reverse()

> The exercise lets you design your own routine for testing the subroutine. It should be tested on several lists, including an empty list. It's important to test it on the empty list since a `null` pointer often represents a special case in an algorithm, and therefor a common source of bugs. It's also a good idea to test a list of length one, for similar reasons. In my `main()` routine, I build up a random list one node at a time and test the `reverse()` subroutine on the list at each step. The `main()` routine was probably harder to write than the subroutine!

> The Solution

>        // This program includes a subroutine that makes a reversed copy of a
>        // list of ints.  The main program simply tests that routine on a few lists.

>        public class ReverseListDemo {

>           static class ListNode {
>                 // Objects of type ListNode are linked together into linked lists.
>              int item;        // An item in the list.
>              ListNode next;   // Pointer to the next node in the list.
>           }

>           static ListNode reverse( ListNode list ) {
>                   // Return a new list containing the same items as the list,
>                   // but in the reverse order.
>              ListNode rev = null;     // rev will be the reversed list.
>              ListNode runner = list;  // For running through the nodes of list.
>              while (runner != null) {
>                     // "Push" the next node of list onto the front of rev.
>                 ListNode newNode = new ListNode();
>                 newNode.item = runner.item;
>                 newNode.next = rev;
>                 rev = newNode;
>                    // Move on to the next node in the list.
>                 runner = runner.next;
>              }
>              return rev;
>           } // end reverse()

>           static void printList(ListNode start) {
>                 // Prints the items in the list to which start points.
>                 // They are printed on one line, separated by spaces.
>               ListNode runner;  // For running along the list.
>               runner = start;
>               while (runner != null) {
>                  System.out.print(" " + runner.item);
>                  runner = runner.next;
>               }
>           } // end printList()

>           static void main(String[] args) {

>              System.out.println("I will print out a list and its reverse, for");
>              System.out.println("several different lists.  The first list is empty.\n");

>              ListNode list = null;   // A list, initially empty.
>              ListNode reversedList;  // The reversed list.

>              int ct = 0;  // How many lists have we processed?

>              while (true) {
>                     // Print the current list and its reverse.  Then
>                     // add a new node onto the head of the list before
>                     // repeating.
>                  System.out.print("The list:          ");
>                  printList(list);
>                  System.out.println();
>                  reversedList = reverse(list);
>                  System.out.print("The reversed list: ");
>                  printList(reversedList);
>                  System.out.println();
>                  System.out.println();
>                  ct++;
>                  if (ct == 6)
>                     break;
>                  ListNode head = new ListNode();  // A new node to add to the list.
>                  head.item = (int)(Math.random()*100);  // A random item.
>                  head.next = list;
>                  list = head;
>              }

>           } // end main()

>        } // end class ReverseListDemo

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