1. 1 CMPT 117 Linked Lists (singly linked)

2. 2 Problems with arrays  When an element is deleted from or inserted into an array, the rest of the array has to be adjusted  If the number of elements is more than what has been allocated, a new array has to be used

3. 3 Linked list  Advantages Easy to insert and delete data Easy to insert and delete data Dynamically allocate data Dynamically allocate data  Disadvantages Random access is slow Random access is slow Need some space overhead to store pointers Need some space overhead to store pointers

4. 4 Pointer based linked list itemnext Can be a pointer to a class Pointer to the next node Each node has two parts:

5. 5 Simple examples using linked list 10

6. 6 Simple examples using linked list 10 30

7. 7 Simple examples using linked list 10 30 103050

8. 8 Insertion 10405060 45

9. 9 Deletion 10452069 Delete this node

10. 10 C++ definition struct Node { PStudent item; Node* next; }; typedef Node* PNode; struct Node { int item; Node* next; }; typedef Node* PNode;  The definition is different for different types of items

11. 11 Allocate space for a node PNode p; p = new Node; p

12. 12 Basic operations  To reference the member Item p->item p->item  To access the next node (if present) p->next p->next  To determine the end of list if (p->next == NULL) if (p->next == NULL)  To access the head of list Needs to define a head pointer, i.e. PNode head Needs to define a head pointer, i.e. PNode head head

13. 13 Class LinkedList  A LinkedList class should be defined to encapsulate all operations of linked list class LinkedList { protected: protected: PNode head; PNode head; public: public: void InsertFirst(...); void InsertFirst(...); void DeleteFirst(...); void DeleteFirst(...); void Insert(...); void Insert(...);......}

14. 14 To delete the first node  Is this code correct? if (head!=NULL) head = head->next;  This code works in Java, but not in C++, because Java has garbage collection  Is this correct? if (head!=NULL) { delete head->item; delete head->item; delete head; delete head; head = head->next; }  This code may work but is dangerous. How should we change it?

15. 15 To insert as the first node  A new node needs to be allocated  The new node should point to the first node of the linked list  head should point to the newly created node  Please write the code for InsertFirst(PStudent pst)?

16. 16 To display the contents of a linked list A high level Pseudocode: 1. Let current pointer point to the first node in the linked list. 2. While (current pointer is not NULL) { Display the data portion of the current node Set current pointer to the Next pointer of the current node }

17. 17 Solution  Use Cur to keep track of the current node and initialise it to Head  To display the data cout item item <<endl;  To advance to the next node Cur = Cur->next;

18. 18 Solution (cont’d) The above can be combined: for (PNode Cur = head; Cur!= NULL; Cur=Cur->next) cout item<<endl;

19. 19 Deleting a node (not the first one) head 58950 CurPre Return this to the system

20. 20 Deleting a node (cont’d)  Locate the node  Adjust the next pointer of the previous node, to point to the next node pointed to by next of the current node Need to have a pointer to point to the previous node! Need to have a pointer to point to the previous node!  Return the deleted node to the system delete or simply collect them all! delete or simply collect them all!

21. 21 To locate a node while( Cur!=NULL && Cur->item!=value) { Pre = Cur; Cur= Cur->next; }

22. 22 To adjust the pointer if (Cur!= NULL) { if (Pre == NULL) { head = Cur->next; //The first node is deleted } else { // not the first node Pre->next = Cur->next; } return Cur; }

23. 23 Insert a node and keep the order of data ascending 2040 PreCur2040 PreCur30 NewPtr Locate the insertion location Adjust the pointers NewPtr->Next=Cur; Pre->Next=NewPtr;

24. 24 Head 20 10 PreCur NewPtr For the first node: NewPtr->Next = Cur; Head = NewPtr; Insert a node (cont’d)

25. 25 To locate a node Pre = NULL; Cur = Head; while(NewValue>Cur->item) { Pre = Cur; Cur = Cur->next; } Will this work? What if NewValue>the value of the last element? Then Cur = Cur->next equals to NULL. So, Cur->item does not exist.

26. 26 To locate the position (revised) Pre= NULL; Cur = Head; while(Cur!= NULL && NewValue>Cur->item) { Pre= Cur; Cur = Cur->next; }