1. Linked List (Part I)

3. Introduction  Weakness of storing an ordered list in array: Insertion and deletion of arbitrary elements are expensive. ○ Example: Given an array which is arranged in ascending order. ○ Discuss how to insert a new element ‘1’ and how to delete the element ‘4’. Storage allocation is not flexible. 2467

4. Possible Improvements  The elements in an ordered list don’t need to be stored in consecutive memory space. The insertion and deletion of an element will not induce excessive data movement.  The element can be “dynamically” allocated.

5. Linked Representation  Data structure for a linked list: first Data Link (pointer): used to store the address of the next node. Node

6. Example BAT3CAT4FAT0 8 first 0 1 2 CAT4 3 FAT0 4 5 6 7 BAT3 8 9 8 first

7. Insertion BAT3CAT4FAT0 8 first 0 1 2 CAT4 3 FAT0 4 5 6 7 BAT3 8 9 8 first  Insert EAT into an ordered linked list 1) Get a new node a 2) Set the data field of a to EAT. 3) Set the link field of a to point the node after CAT, which contains FAT. Find the position where EAT is to be inserted. EAT 6 6 4 6 4CAT6 6 3 4) Set the link field of the node containing CAT to a.

8. Deletion BAT3CAT6EAT4 8 first 1 2 CAT6 3 FAT0 4 5 EAT4 6 7 BAT3 8 9 8 first  Remove CAT from the linked list 1) Set the link of BAT to EAT. 2) Deallocate CAT Find the address of CAT FAT0 BAT6 8 6 3

9. Representation of a Linked List class ListNode { friend class LinkedList; public: ListNode(); ListNode(DataField value); ~ListNode(); private: DataField data; ListNode *link; }; class LinkedList { private: ListNode * first; }; first class LinkedList { private: ListNode * first; }; class ListNode { friend class LinkedList; public: ListNode(); ListNode(DataField value); ~ListNode(); private: DataField data; ListNode *link; }; datalink

11. Linked Ordered List  Suppose elements are arranged in ascending order.  ADT class LinkedOrderedList { public: LinkedOrderedList(); ~ LinkedOrderedList(); void Insert(DataField value); bool Delete(DataField value); //return false if value is not found. bool IsEmpty(); private: ListNode *first; };

12. Initialization  The constructor of ListNode:  The constructor of LinkedOrderedList: LinkedOrderList::LinkedOrderList() { first = NULL; } ListNode::ListNode(DataField value) { data = value; link = NULL; }

13. Algorithm of Insert() 01void LinkedOrderList::Insert(DataField value) 02{ 03 curr = first; 04 while (curr != NULL) 05 { 06if (curr->data >= value) 07 { 08 ListNode *a = new ListNode(value); 09 a->link = curr; 10 previous->link = a; 11 break; 12 } 13 previous = curr; 14 curr = curr->link; 15 } 16}

14. Insertion BAT3CAT4FAT0 8 first  Insert EAT into an ordered linked list curr EAT 4CAT6 previous 03 curr = first; 04 while (curr != NULL) 05 { 06 if (curr->data >= value) 07 { 08 ListNode *a = new ListNode(value); 09 a->link = curr; 10 previous->link = a; 11 break; 12 } 13 previous = curr; 14 curr = curr->link; 15 } a

15. Boundary Condition: Case 1  Consider to insert AT. There will be no previous node for AT. The update of first is required. BATCATFAT first AT

16. Boundary Condition: Case 2  Consider to insert GAT. BATCATFAT first GAT 03 curr = first; 04 while (curr != NULL) 05 { 06 if (curr->data >= value) 07 { 08 ListNode *a = new ListNode(value); 09 a->link = curr; 10 previous->link = a; 11 break; 12 } 13 previous = curr; 14 curr = curr->link; 15 } No statement is written to insert GAT at the end of the list.

17. Problem of Insert()  The function Insert() fails to deal with boundary conditions. The insertion is always performed between two existing nodes.  Improvements Add codes before- and after the while-statement for dealing with the boundary conditions. Always maintain two (dummy) nodes so that insertion can always be performed between two nodes.

18. Improvement Using Two Dummy Nodes  Maintain two dummy nodes at each end of the list. class LinkedList { private: ListNode * first, *last; }; LinkedOrderList::LinkedOrderList() { first = new ListNode(); last = new ListNode(); first->link = last; last->link = NULL; } firstlast  No need to update the pointer first.  Boundary conditions are eliminated (Insertion and Deletion always take place between two nodes).

19. New Version of Insert() 01void LinkedOrderList::Insert(DataField value) 02{ 03 previous = first; 04 curr = first->link; 05 while (curr != NULL) 06 { 07if (curr == last || curr->data >= value) 08 { 09 ListNode *a = new ListNode(value); 10 a->link = curr; 11 previous->link = a; 12 break; 13 } 14 previous = curr; 15 curr = curr->link; 16 } 17}

20. Algorithm of Delete() 01bool LinkedOrderList::Delete(DataField value) 02{ 03 if (IsEmpty()) 04 return false; 05 06 previous = first; 07 curr = first->link; 08 while (curr != last) 09 { 10 if (curr->data == value) 11 { 12 previous->link = curr->link; 13 Deallocate curr; 14 return true; 15 } 16 previous = curr; 17 curr = curr->link; 18 } 19 return false; 20}

21. Deletion  Consider to remove BAT from the list. firstlast BAT 10 if (curr->data == value) 11 { 12 previous->link = curr->link; 13 Deallocate curr; 14 return true; 15 } previouscurrcurr->link

22. Destruction of Nodes  Remember to deallocate each node in the destructor. 01 LinkedOrderList ::~ LinkedOrderList() 02{ 03 curr = first; 04 while (curr != NULL) 05 { 06 next = curr->link; 07 Deallocate curr; 08 curr = next; 09 } 10}

23. Performance Analysis  Suppose there are n nodes in a linked list. Space complexity: ○ A linked list uses an exact amount of memory space to store these n nodes. ○ Space complexity to perform a insertion or deletion O(1). Time complexity to perform a insertion or deletion: ○ Consider a worst case in which nodes are always inserted and deleted at the end of the list. Therefore, the complexity is O(n). ○ Excessive request of allocation or deallocation of memory space for a node increases loading for the OS system (and may lower efficiency).

25. Linked Stack BACE0 datalink top Pop D Push D

26. Linked Queue  Deletion takes place at front; insertion at rear. BCDAE0 frontrear Pop Push E

27. Implementation of Linked Queue LinkedQueue:: LinkedQueue() { front = rear = NULL; }; bool LinkedQueue::IsEmpty() { if (front is NULL and rear is NULL) return true; return false; }

28. Implementation of Linked Queue void LinkedQueue::Push(Datafield value) { if (IsEmpty()) front = rear = new ListNode(value); else rear = rear->link = new ListNode(value); }; Datafield LinkedQueue::Pop() { if (IsEmpty()) output error; else { delNode = front; value = front->data; front = front->link; deallocate delNode; return value; } }; LinkedQueue::~ LinkedQueue() { while (!IsEmpty()) Pop(); };

29. Comparison  Compare stack/queue implemented using array and linked stack/queue. ArrayLinked list Memory spaceThe length of an array is fixed; Resize() is required if the stack is full. Memory space can be dynamically allocated. The storage is more compact. Execution time for Push() and Pop() The time complexity is O(1). The time complexity is also O(1). But the memory request increase overhead.

31. Polynomial Representation class Polynomial { private: ListNode *first; }; first class ListNode { friend class LinkedList; public: ListNode(int c, int e); ~ListNode(); private: int coef, exp; ListNode *link; }; 3210 f(x) =3x 2 +1 coefexp

32. Adding Polynomial  Example Consider the following two polynomials: a(x) =3x 14 +2x 8 +1 b(x) =8x 14 -3x 10 +10x 6 b.first 814-310 bi a.first 31428 ai 1 0 010 0 6

33. Case 1  ai->exp == bi->exp 814-310 bi 31428 ai 1 0 0 C.first 11 0 14 Add coefficients and append to the result C. Advance ai and bi to next term. 10 0 6

34. Case 2  ai->exp exp 814-310 ai 31428 bi 1 0 0 C.first -3 0 10 Append the term indicated by bi to the result C. Advance bi to next term. 111410 0 6

35. Case 3  ai->exp > bi->exp 814-310 bi 31428 ai 1 0 0 C.first -310 Append the term indicated by ai to the result C. Advance ai to next term. 111410 0 628 61 0 0

36. Algorithm of Add() LinkedPolynomial LinkedPolynomial::Add(Polynomial B) { Create a new LinkedPolynomial C as result; ai = first; bi = B.first; while (ai != NULL && bi != NULL) { if (ai->exp == bi->exp) { Sum = ai->coef + bi->coef; if (Sum != 0) C.InsertBack(Sum, ai->exp); ai = ai->link; bi = bi->link; } else if (ai->exp exp) { C.InsertBack(bi->coef, bi->exp); bi = bi->link; } else if (ai->exp > bi->exp) { C.InsertBack(ai->coef, ai->exp); ai = ai->link; } for (; ai != NULL; ai = ai->link) C.InsertBack(ai->coef, ai->exp); for (; bi != NULL; bi = bi->link) C.InsertBack(bi->coef, bi->exp); return C; }