CS 136 - Lecture 10

1. Extending Classes: Inheritance
2. Other linked list representations
1. Circularly-linked lists

Extending Classes: Inheritance

A can also add new methods or redefine the "inherited" methods.

When redefine methods, can call original version from superclass:

   super.meth(a,b)

   super(...)

See examples with Position and RowOrderPosn as well as CurDoublyLinkedList as extension of DoublyLinkedList.

Linked List Implementations (cont'd.)

Circularly Linked Lists

• Notice that tailPeek, addToTail, and removeFromTail are the only operations that are more expensive in the linked representation than Vector.

• The problem is that it takes O(n) time to find the end. We can overcome this by adding a new field to the class: tail, which should serve as a pointer to the tail of the list.

• An empty list has head=tail=null, a list w/one elt has head = tail = reference to that elt, while in all other cases, head != tail.

• What is the cost of this? Adds complexity to add and remove methods since must worry about resetting tail field. Even addToHead may have to reset tail field (why?).

• It also doesn't help with removeFromTail since need predecessor!

• Can simplify further by noticing that tail node of list has wasted next field = null.

• Why not use that field to point to beginning of list, then don't need head field?

public class CircularList implements List {

   protected SinglyLinkedListElement tail;
   protected int count;

   public CircularList()
   // pre: constructs a new circular list
   {
      tail = null;
      count = 0;
   }

   public void add(Object value)
   // post: adds value to beginning of list.
   {
      addToHead(value);
   }

   public void addToHead(Object value)
   // pre: value non-null
   // post: adds element to head of list
   {
      SinglyLinkedListElement temp =
          new SinglyLinkedListElement(value);
      if (tail == null) {
          tail = temp;
          tail.setNext(tail);
      }
      else {
          temp.setNext(tail.next());
          tail.setNext(temp);
      }
      count++;
   }

   public void addToTail(Object value)
   // pre: value non-null
   // post: adds element to tail of list
   {
      addToHead(value);
      tail = tail.next();      // moves new from head to tail
   }

   public Object peek()
   // pre: !isEmpty()
   // post: returns value at head of list
   {
      return tail.next().value();
   }

   public Object tailPeek()
   // pre: !isEmpty()
   // post: returns value at tail of list
   {
      return tail.value();
   }

   public Object removeFromHead()
   // pre: !isEmpty()
   // post: returns and removes value from head of list
   {
      SinglyLinkedListElement temp = tail.next();  // ie. the head of the list

      if (tail == tail.next())                     // 1 elt in list
          tail = null;
      else {
          tail.setNext(temp.next());
          temp.setNext(null);             // helps clean things up
      }              // temp is free
      count--;
      return temp.value();
   }

   public Object removeFromTail()
   // pre: !isEmpty()
   // post: returns and removes value from tail of list
   {
      Assert.pre(!isEmpty(),"The list is not empty.");
      SinglyLinkedListElement finger = tail;
      while (finger.next() != tail)
          finger = finger.next();
      // finger now points to second-to-last value
      SinglyLinkedListElement temp = tail;
      if (finger == tail)
          tail = null;
      else {
          finger.setNext(tail.next());
          tail = finger;
      }
      count--;
      return temp.value();
   }

   public boolean contains(Object value)
   // pre: value != null
   // post: returns true if list contains value, else false
   {
      if (tail == null) return false;

      SinglyLinkedListElement finger;
      finger = tail.next();
      while ((finger != tail) && (!finger.value().equals(value)))
          finger = finger.next();
      return finger.value().equals(value);
   }

   public Object remove(Object value)
   // pre: value != null
   // post: remove & returns element equal to value, or null
   {
      if (tail == null) return null;

      SinglyLinkedListElement finger = tail.next();
      SinglyLinkedListElement previous = tail;
      int compares;
      for (compares = 0;
           (compares < count) && (!finger.value().equals(value));
           compares++)
      {
          previous = finger;
          finger = finger.next();
      }
      if (finger.value().equals(value)) {
          // an example of the pigeon-hole principle
          if (tail == tail.next())
              tail = null;
          else {
              if (finger == tail)
                  tail = previous;
              previous.setNext(previous.next().next());
          }
          // finger value free
          finger.setNext(null)       // to keep things disconnected
          count--;                   // fewer elements
          return finger.value();
      }
      else
      return null;
   }

   public int size()
   // post: returns number of elements in list
   {
      return count;
   }

   public boolean isEmpty()
   // post: returns true if no elements in list
   {
      return tail == null;
   }

   public void clear()
   // post: removes all elements from list.
   {
      count = 0;
      tail = null;
   }
}

• With this implementation, addToTail is now O(1).

• But now it takes one extra dereference (following reference) to get to head.

• Only contains, remove, and removeFromTail are still O(n).

• Contains and remove involve searches and seem likely always to be O(n) (unless we attempt to keep list in order and do binary search - which has its own problems - come back to it later!).

• However, why can't we make removeFromTail O(1)? The problem is that we need to know the predecessor in order to delete an element from the list.

• Unfortunately references only go from the front to the back of the list. Why not put them in the other direction instead? Then it would be harder to delete from the front.