// node of generic doubly linked list class

public class DLLNode { public T info; public DLLNode next, prev; public DLLNode() { next = null; prev = null; } public DLLNode(T el) { info = el; next = null; prev = null; } public DLLNode(T el, DLLNode n, DLLNode p) { info = el; next = n; prev = p; } }

//**************************** DLL.java ******************************* // generic doubly linked list class

public class DLL { private DLLNode head, tail; public DLL() { head = tail = null; } public boolean isEmpty() { return head == null; } public void setToNull() { head = tail = null; } public T firstEl() { if (head != null) return head.info; else return null; } public void addToHead(T el) { if (head != null) { head = new DLLNode(el,head,null); head.next.prev = head; } else head = tail = new DLLNode(el); } public void addToTail(T el) { if (tail != null) { tail = new DLLNode(el,null,tail); tail.prev.next = tail; } else head = tail = new DLLNode(el); } public T deleteFromHead() { if (isEmpty()) return null; T el = head.info; if (head == tail) // if only one node on the list; head = tail = null; else { // if more than one node in the list; head = head.next; head.prev = null; } return el; } public T deleteFromTail() { if (isEmpty()) return null; T el = tail.info; if (head == tail) // if only one node on the list; head = tail = null; else { // if more than one node in the list; tail = tail.prev; tail.next = null; } return el; } public void printAll() { for (DLLNode tmp = head; tmp != null; tmp = tmp.next) System.out.print(tmp.info + " "); } public T find(T el) { DLLNode tmp; for (tmp = head; tmp != null && !tmp.info.equals(el); tmp = tmp.next); if (tmp == null) return null; else return tmp.info; } }

public class DLLTest {

public static void main(String[] args) {

DLL test = new DLL();

for(int i = 0; i

test.addToTail("a" + i);

test.printAll();

}

}

Question IV) Doubly circular linked lists (10 pts): Assume that a circular doubly linked list has been created, as in the figure below. After each of the following assignments, indicate changes made in the list by showing which links have been modified. Process these assignments in sequence; that is, the second assignment should make changes in the list modified by the first assignment, and soon list.next.next.next-list.prev; list.prev.prev.prev - list.next.next.next.prev; list.next.next.next.prev- list.prev.prev.prev; list.next-list.next.next; list.next.prev.next- list.next.next.next; list