1. Linked Lists
CSM Linked Lists

2. ‘Variable-length arrays’?
Arrays are fixed-length
To expand them, you create a new, longer array, and copy the contents of the old array into the new array
This is slow!
Solution: Attach a pointer to each item in the array, which points to the next item
This is a linked list
An data item plus its pointer is called a node
CSM Linked Lists

3. The Linked List data structure
[0]
[1]
[2]
array A B C
Array
node
linked A B C
Linked list
Linked lists are unbounded
(maximum number of items limited only by memory)
CSM Linked Lists

4. The Node data structure
Contains two things:
A piece of data (any object)
The next node in the list
Write the code for a node…
CSM Linked Lists

5. The Linked List data structure
Same operations as for arrays. Here are the ones we know:
addToFront, addToBack: add an object to the front/back of the list
removeFromFront/Back: remove an object from the front/back
getFront/Back: examine the object at the front/back (no removal)
isEmpty: determines whether or not the list is empty
length: returns the number of objects in the list
equals: tests two lists for equality
toString: converts a list to a String
'Ordered' means that the order matters
CSM Linked Lists

6. linked list Some common examples of a linked list
Hash tables use linked lists for collision resolution
Any "File Requester" dialog uses a linked list
Binary Trees
Stacks and Queues can be implemented with a doubly linked list
Relational Databases (eg. Microsoft Access)
Exercise: method signatures for the class (see previous slide)
CSM Linked Lists

7. Intialization Node head, Tail; Head=tail=null; class Node { int data;
Node next; }

8. 2. Adding a Node Adding an element to the head of a linked list
public void CreateNode (int element) {
Node current=new Node (element);
if (head==null)
head=current;
tail=current; }
else
current.next=head;

9. Head == null
Head
Tail

10. Head == null
Current
Head A
Tail
Node current=new Node (element);

11. head=current;
tail=current;
Head A
Tail

12. if (head !=null)
Node current=new Node (B);
Head B A
current
Tail

13. current.next=head;
head=current;
Head B A
Tail

14. current.next=head;
head=current;
Head C B A
current
Tail

15. Adding an element to the tail
public void CreateNode(int element) {
Node current=new Node(element);
current.next=null;
if (head==null)
head=current;
tail=current; }
else
tail.next=current;

16. tail.next=current;
tail=current;
Head A B
current
Tail

17. tail.next=current;
tail=current;
Head A B
Tail
Tail

18. removeFromHead head=head.next; Initial list Remove Head (C) Head C B A
Tail
Remove Head (C)
Implementation is left as an exercise. Remember to return the removed object.
head=head.next;
Head B A
Tail
CSM Linked Lists

19. Removing an element from a linked list
public int DeleteNode(int position) {
int ptr=1;
if (position==1) {
head=head.next;
return(1); }
CSM Linked Lists

20. current=head;
while (current.next!=null) {
if(ptr==position-1)
break;
current=current.next;
ptr++; }
if (ptr==position-1) {
current.next=current.next.next;
return (1);
else return (-1);

21. Consider the polynomial expression: A(x)= amxem+…….+a1xe1 ,
Polynomial Addition
Consider the polynomial expression:
A(x)= amxem+…….+a1xe1 ,
where ai are non-zero coefficients and ei are
exponents

22. The polynomial can be represented using linked list.
Each term in the expression will be represented by a node.
A node will be of fixed size having 3 fields which represent the coefficient and exponent of a term plus a pointer to the next term.
Coeff
Exp
Next

23. For eg: A= 3x14 + 2x8 +1 would be stored as

24. We have two polynomial.
A= 3x14 + 2x8 +1
B= 8x14 – 3x x6
Their sum will be
11x14 – 3x10 + 2x8 +10x6 + 1.

25. We have two pointers p and q that points to A and B respectively.
if the exponents of the two terms are equal , then the coefficients are added and a new term is created for the result.
If the exponents of the current term in A is less than the current term in B, then a duplicate of B is created and attached to C. the pointer q is advanced to the next term.
Similar action is taken on A if exp(p) > exp(q).

26. Each time a new node is generated its coeff and exp fields are set and it is appened to the end of the list C.
d is a pointer that points to the last node in the resultant node.

27. Procedure ATTACH(c,e,d)
//c – coeff, e – exp, d – pointer
call GETNODE(I)
exp(I)  e
coeff (I) c
link(d)  I
d  I
end ATTACH

29. procedure PADD(A,B,C)
// A and B – singly linked list are summed to form the new
// list name C
p  A
q  B
call GETNODE (C)
d  C
while p  0 and q  0 do

30. case : exp (p) = exp(q)
x  coeff(p) + coeff(q)
if x  0 then
call ATTACH( x, exp(p), d)
p  link(p)
q  link(q)

31. case : exp (p) < exp (q) call ATTACH( coeff (q), exp(q), d)
q  link(q)
else
call ATTACH( coeff (p), exp(p), d)
p  link(p)
end case
endwhile

32. while p  0 do
call ATTACH( coeff(p), exp(p), d)
p  link(p)
endwhile
while q  0 do
call ATTACH( coeff(q), exp(q), d)
q  link(q)
link(d)  0;t  C ; C link(C)
call RET(T)
end PADD

33. Thank you……