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Chapter 5 – Dynamic Data Structure Part 2: Linked List DATA STRUCTURES & ALGORITHMS Teacher: Nguyen Do Thai Nguyen

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Presentation on theme: "Chapter 5 – Dynamic Data Structure Part 2: Linked List DATA STRUCTURES & ALGORITHMS Teacher: Nguyen Do Thai Nguyen"— Presentation transcript:

1 Chapter 5 – Dynamic Data Structure Part 2: Linked List DATA STRUCTURES & ALGORITHMS Teacher: Nguyen Do Thai Nguyen E-mail: nguyenndt@hcmup.edu.vnnguyenndt@hcmup.edu.vn Phone: 0902578939 Group: https://www.facebook.com/groups/335118686690998 Ho Chi Minh City University of Pedagogy

2 Outline Linked List 5. Array versus Linked Lists 1. Concept of Linked List 2. Declaration of Linked List 3. Operation of Linked List 4. Variations of Linked Lists

3 Problem With Array Fixed Length (Occupy a Block of memory) [Are also called Dense List] – > waste space Insertion or Deletion: slow 3

4 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 4

5 1. Concept of Linked List * A linked list, or one-way list, is a linear collection of data elements, called nodes, where the linear order is given by means of pointer 5

6 * Each node is divided into two parts n First part contains the information of the element, and n Second part called the link field or nextpointer field contains the address of the next node in the list. 6 1. Concept of Linked List

7 * A linked list is a series of connected nodes * Head : pointer to the first node  The last node points to NULL 7 Head  A BC

8 1. Concept of Linked List * Each node contains at least n A piece of data (any type) n Pointer to the next node in the list 8 datapointer node A

9 2. Declaration of Linked List  We use two classes: Node and List  Declare Node class for the nodes data : double -type data in this example next : a pointer to the next node in the list class Node { public: doubledata; // data Node*next; // pointer to next };

10 2. Declaration of Linked List  Declare List, which contains head : a pointer to the first node in the list. Since the list is empty initially, head is set to NULL Operations on List

11 2. Declaration of Linked List class List { public: List(void) { head = NULL; }// constructor ~List(void);// destructor bool IsEmpty() { return head == NULL; } Node* InsertNode(int index, double x); int FindNode(double x); int DeleteNode(double x); void DisplayList(void); private: Node* head; };

12 3. Operation of Linked List * Operations of List n IsEmpty: determine whether or not the list is empty n InsertNode: insert a new node at a particular position n FindNode: find a node with a given value n DeleteNode: delete a node with a given value n DisplayList: print all the nodes in the list

13 Inserting a new node * Node* InsertNode(int index, double x) n Insert a node with data equal to x after the index’th elements. (i.e., when index = 0, insert the node as the first element; when index = 1, insert the node after the first element, and so on) n If the insertion is successful, return the inserted node. Otherwise, return NULL. (If index is length of the list, the insertion will fail.)

14 Inserting a new node * Steps 1. Locate index’th element 2. Allocate memory for the new node 3. Point the new node to its successor 4. Point the new node’s predecessor to the new node newNode index’th element

15 Inserting a new node  Possible cases of InsertNode 1. Insert into an empty list 2. Insert in front 3. Insert at back 4. Insert in middle * But, in fact, only need to handle two cases n Insert as the first node (Case 1 and Case 2) n Insert in the middle or at the end of the list (Case 3 and Case 4)

16 Inserting a new node Node* Node* List::InsertNode(int index, double x) { if (index < 0) return NULL; int currIndex=1; Node* currNode=head; while (currNode && index > currIndex) { currNode=currNode->next; currIndex++; } if (index > 0 && currNode == NULL) return NULL; Node* newNode=newNode; newNode->data=x; if (index == 0) { newNode->next=head; head=newNode; } else { newNode->next=currNode->next; currNode->next=newNode; } return newNode; } Try to locate index’th node. If it doesn’t exist, return NULL.

17 Inserting a new node Node* Node* List::InsertNode(int index, double x) { if (index < 0) return NULL; int currIndex=1; Node* currNode=head; while (currNode && index > currIndex) { currNode=currNode->next; currIndex++; } if (index > 0 && currNode == NULL) return NULL; Node* newNode=newNode; newNode->data=x; if (index == 0) { newNode->next=head; head=newNode; } else { newNode->next=currNode->next; currNode->next=newNode; } return newNode; } Create a new node

18 Inserting a new node Node* Node* List::InsertNode(int index, double x) { if (index < 0) return NULL; int currIndex=1; Node* currNode=head; while (currNode && index > currIndex) { currNode=currNode->next; currIndex++; } if (index > 0 && currNode == NULL) return NULL; Node* newNode=newNode; newNode->data=x; if (index == 0) { newNode->next=head; head=newNode; } else { newNode->next=currNode->next; currNode->next=newNode; } return newNode; } Insert as first element head newNode

19 Inserting a new node Insert after currNode Node* Node* List::InsertNode(int index, double x) { if (index < 0) return NULL; int currIndex=1; Node* currNode=head; while (currNode && index > currIndex) { currNode=currNode->next; currIndex++; } if (index > 0 && currNode == NULL) return NULL; Node* newNode=newNode; newNode->data=x; if (index == 0) { newNode->next=head; head=newNode; } else { newNode->next=currNode->next; currNode->next=newNode; } return newNode; } newNode currNode

20 Finding a node * int FindNode(double x) Search for a node with the value equal to x in the list. n If such a node is found, return its position. Otherwise, return 0. int List::FindNode(double x) { Node* currNode=head; int currIndex=1; while (currNode && currNode->data != x) { currNode=currNode->next; currIndex++; } if (currNode) return currIndex; return 0; }

21 Finding a node * int FindNode(double x) Search for a node with the value equal to x in the list. n If such a node is found, return its position. Otherwise, return 0. int List::FindNode(double x) { Node* currNode = head; int currIndex=1; while (currNode && currNode->data != x) { currNode=currNode->next; currIndex++; } if (currNode) return currIndex; return 0; }

22 Deleting a node * int DeleteNode(double x) n Delete a node with the value equal to x from the list. n If such a node is found, return its position. Otherwise, return 0. * Steps n Find the desirable node (similar to FindNode ) n Release the memory occupied by the found node n Set the pointer of the predecessor of the found node to the successor of the found node * Like InsertNode, there are two special cases n Delete first node n Delete the node in middle or at the end of the list

23 Deleting a node * int DeleteNode(double x) n Delete a node with the value equal to x from the list. n If such a node is found, return its position. Otherwise, return 0. * Steps n Find the desirable node (similar to FindNode ) n Release the memory occupied by the found node n Set the pointer of the predecessor of the found node to the successor of the found node * Like InsertNode, there are two special cases n Delete first node n Delete the node in middle or at the end of the list

24 Deleting a node * int DeleteNode(double x) n Delete a node with the value equal to x from the list. n If such a node is found, return its position. Otherwise, return 0. * Steps n Find the desirable node (similar to FindNode ) n Release the memory occupied by the found node n Set the pointer of the predecessor of the found node to the successor of the found node * Like InsertNode, there are two special cases n Delete first node n Delete the node in middle or at the end of the list

25 Deleting a node int List::DeleteNode(double x) { Node* prevNode=NULL; Node* currNode=head; int currIndex=1; while (currNode && currNode->data != x) { prevNode=currNode; currNode=currNode->next; currIndex++; } if (currNode) { if (prevNode) { prevNode->next=currNode->next; delete currNode; } else { head=currNode->next; delete currNode; } return currIndex; } return 0; } Try to find the node with its value equal to x

26 Deleting a node int List::DeleteNode(double x) currNode prevNode A B C

27 Deleting a node int List::DeleteNode(double x) prevNode AC

28 Deleting a node int List::DeleteNode(double x) { prevNode AC

29 Deleting a node int List::DeleteNode(double x) { Node* prevNode=NULL; Node* currNode=head; int currIndex=1; while (currNode && currNode->data != x) { prevNode=currNode; currNode=currNode->next; currIndex++; } if (currNode) { if (prevNode) { prevNode->next=currNode->next; delete currNode; } else { head=currNode->next; delete currNode; } return currIndex; } return 0; }

30 Deleting a node int List::DeleteNode(double x) { currNode head AB

31 Deleting a node int List::DeleteNode(double x) { head B

32 Deleting a node int List::DeleteNode(double x) { head B

33 Deleting a node int List::DeleteNode(double x) { Node* prevNode=NULL; Node* currNode=head; int currIndex=1; while (currNode && currNode->data != x) { prevNode=currNode; currNode=currNode->next; currIndex++; } if (currNode) { if (prevNode) { prevNode->next=currNode->next; delete currNode; } else { head=currNode->next; delete currNode; } return currIndex; } return 0; }

34 Printing all the elements * void DisplayList(void) n Print the data of all the elements n Print the number of the nodes in the list void List::DisplayList() { int num=0; Node* currNode=head; while (currNode != NULL) { cout data << endl; currNode=currNode->next; num++; } cout << "Number of nodes in the list: " << num << endl; }

35 Destroying the list * ~List(void) n Use the destructor to release all the memory used by the list. n Step through the list and delete each node one by one. void List::~List(void) { Node* currNode = head, *nextNode = NULL; while (currNode != NULL) { nextNode=currNode->next; // destroy the current node delete currNode; currNode=nextNode; }

36 Using List int main(void) { List list; list.InsertNode(0, 7.0);// successful list.InsertNode(1, 5.0);// successful list.InsertNode(-1, 5.0);// unsuccessful list.InsertNode(0, 6.0);// successful list.InsertNode(8, 4.0);// unsuccessful // print all the elements list.DisplayList(); if(list.FindNode(5.0) > 0)cout << "5.0 found" << endl; elsecout << "5.0 not found" << endl; if(list.FindNode(4.5) > 0) cout << "4.5 found" << endl; elsecout << "4.5 not found" << endl; list.DeleteNode(7.0); list.DisplayList(); return 0; } 6 7 5 Number of nodes in the list: 3 5.0 found 4.5 not found 6 5 Number of nodes in the list: 2 result

37 4. Variations of Linked Lists * Circular linked lists n The last node points to the first node of the list n How do we know when we have finished traversing the list? (Tip: check if the pointer of the current node is equal to the head.) A Head BC

38 4. Variations of Linked Lists * Doubly linked lists n Each node points to not only successor but the predecessor There are two NULL: at the first and last nodes in the list n Advantage: given a node, it is easy to visit its predecessor. Convenient to traverse lists backwards A Head B  C 

39 5. Array versus Linked Lists * Linked lists are more complex to code and manage than arrays, but they have some distinct advantages. n Dynamic: a linked list can easily grow and shrink in size.  We don’t need to know how many nodes will be in the list. They are created in memory as needed.  In contrast, the size of a C++ array is fixed at compilation time. n Easy and fast insertions and deletions  To insert or delete an element in an array, we need to copy to temporary variables to make room for new elements or close the gap caused by deleted elements.  With a linked list, no need to move other nodes. Only need to reset some pointers.

40 Example: The Polynomial ADT * An ADT for single-variable polynomials * Array implementation

41 The Polynomial ADT… n Acceptable if most of the coefficients A j are nonzero, undesirable if this is not the case n E.g. multiply  most of the time is spent multiplying zeros and stepping through nonexistent parts of the input polynomials * Implementation using a singly linked list n Each term is contained in one cell, and the cells are sorted in decreasing order of exponents

42 Overview of Templates  template keyword tells the compiler that the class definition that follows will manipulate one or more unspecified types. template class List * When a list is declared, a specific type must be specified. The compiler then generates the actual class code from the template. int main(void) { // use integer type instead of double List list;... }

43 Linked List using template  Node with/without using template class Node { public: doubledata; Node*next; }; template class Node { public: Objectdata; Node*next; }; * template Object is the substitution parameter, representing a type name. Object is used everywhere in the class where you would normally see the specific type (e.g. double ) the container holds.

44 Linked List using template template class List { public: List(void) { head = NULL; } ~List(void); bool IsEmpty() { return head == NULL; } Node * InsertNode(int index, Object & x); int FindNode(Object & x); int DeleteNode(Object & x); void DisplayList(void); private: Node * head; };

45 Linked List using template The first line indicates that Object is the template argument template List ::~List(void) { // desctructor Node * currNode = head, *nextNode = NULL; while (currNode != NULL) { nextNode = currNode->next; delete currNode; currNode = nextNode; }

46 template Node * List ::InsertNode(int index, Object & x) { if (index < 0) return NULL; int currIndex=1; Node * currNode=head; while (currNode && index > currIndex) { currNode=currNode->next; currIndex++; } if (index > 0 && currNode == NULL) return NULL; Node * newNode=newNode ; newNode->data=x; if (index == 0) { newNode->next=head; head=newNode; } else { newNode->next=currNode->next; currNode->next=newNode; } return newNode; }

47 Linked List using template template int List ::FindNode(Object & x) { Node * currNode=head; int currIndex=1; while (currNode && currNode->data != x) { currNode=currNode->next; currIndex++; } if (currNode) return currIndex; return 0; }

48 template int List ::DeleteNode(Object & x) { Node * prevNode=NULL; Node * currNode=head; int currIndex=1; while (currNode && currNode->data != x) { prevNode= currNode; currNode= currNode->next; currIndex++; } if (currNode) { if (prevNode) { prevNode->next= currNode->next; delete currNode; } else { head= currNode->next; delete currNode; } return currIndex; } return 0; }

49 Linked List using template template void List ::DisplayList() { int num=0; Node * currNode=head; while (currNode != NULL) { cout data << endl; currNode=currNode->next; num++; } cout << "Number of nodes in the list: " << num << endl; }

50 Using List int main(void) { // use integer type instead of double List list; int x1 = 7, x2 = 5, x3 = 6, x4 = 8; // add items to the list list.InsertNode(0, x1); list.InsertNode(0, x2); list.InsertNode(0, x3); list.InsertNode(1, x4); // print all the elements list.DisplayList(); if(list.FindNode(x2) > 0)cout << "5 found" << endl; elsecout << "5 not found" << endl; list.DeleteNode(x1); list.DisplayList(); return 0; }

51

52 Execricise 1.Merge two ordered single linked lists of intergers into one ordered list. 2.Delete an ith node on a linked list. Be sure that such a node exists 3.Delete from list L1 nodes whose positions are to be found in an ordered list L2. For instance, if L1 = (A B C D E) and L2= (2 4 8), then the second and the fourth nodes are to be deleted from list L1 (the eighth node does not exist), and after deletion, L1 = (A C E) 52

53 Execricise 4.Delete from list L1 nodes occupying position indeicated in ordered lists L2 and L3. For instance, if L1 = (A B C D E), L2 = (2 4 8), and L3 = (2 5), then after deletion, L1 = (A C) 5.Delete from an ordered list L nodes occupying positions indicated in list L itself. For instance, if L = (1 3 5 7 8), then after deletion, L = (1 7) 53

54 Execricise 6. Write a member function to check whether two singly linked lists have the same contents 7. Write a member function to reverse a singly linked list using only one pass through the list 8. Creat a singly linked list that is displayed in the picture below: 54

55 Execricise 9. Make Menu and perform the following operation by the help of singly linked list?linked list – Create a link list. (The list should be Dynamic and should stop only when the user wishes to end the creation of the list) – Ask User to enter a desire node and as a result the program will delete that node from the list. – Take a node from the user as input and insert that node after the desired node provided by the user.node – Take an identification of a node as input from the user and search that node from the created link list.node search 55

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