Linked Lists Geletaw S Objective At the end
Linked Lists Geletaw S.
Objective At the end of the session students should have basic understanding of – Linked lists – Basic operations of linked lists • Insert, find, delete, print, etc. – Variations of linked lists • Simple Linked Lists • Circular Linked Lists • Doubly Linked Lists
Array vs Linked List Array Linked List node
What’s wrong with Array ? Disadvantages of arrays as storage data structures – slow searching in unordered array – slow insertion in ordered array – Fixed size
Why lists (Linked lists )? • Solve some of these problems • General purpose storage data structures and are versatile. • More complex to code and manage than arrays, but they have some distinct advantages. • 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. • 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.
Linked Lists • Each data item is embedded in a link. • Each Link object contains a reference to the next link in the list of items. In an array – items have a particular position, identified by its index. In a list – the only way to access an item is to traverse the list
Linked Lists A B C Head • A linked list is a series of connected nodes • Each node contains at least – A piece of data (any type) – Pointer to the next node in the list • Head: pointer to the first node • The last node points to NULL node A data pointer
A Simple Linked List Class Operations of List Is. Empty: – determine whether or not the list is empty Insert. Node: – insert a new node at a particular position Find. Node: – find a node with a given value Delete. Node: – delete a node with a given value Display. List: – print all the nodes in the list
A Simple Linked List Class • 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: double data; // data Node* next; // pointer to next };
A Simple Linked List Class • 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 class List { public: List(void) { head = NULL; } ~List(void); // constructor // destructor bool Is. Empty() { return head == NULL; } Node* Insert. Node(int index, double x); int Find. Node(double x); int Delete. Node(double x); void Display. List(void); private: Node* head; };
Inserting a new node • Node* Insert. Node(int index, double x) – 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) – If the insertion is successful, return the inserted node. – Otherwise, return NULL. (If index is < 0 or > length of the list, the insertion will fail. ) Steps 1. 2. 3. 4. Locate index’th element Allocate memory for the new node Point the new node to its successor Point the new node’s predecessor to the new node index’th element new. Node
Inserting a new node Possible cases of Insert. Node 1. 2. 3. 4. Insert into an empty list Insert in front Insert at back Insert in middle But, in fact, only need to handle two cases – Insert as the first node (Case 1 and Case 2) – Insert in the middle or at the end of the list (Case 3 and Case 4)
Inserting a new node Node* List: : Insert. Node(int index, double x) { if (index < 0) return NULL; Try to locate index’th node. If it doesn’t exist, return NULL. int curr. Index = 1; Node* curr. Node = head; while (curr. Node && index > curr. Index) { curr. Node = curr. Node->next; curr. Index++; } if (index > 0 && curr. Node == NULL) return NULL; Node* new. Node = new. Node->data = if (index == 0) { new. Node->next head } else { new. Node->next curr. Node->next } return new. Node; } new x; Node; = = head; new. Node; = = curr. Node->next; new. Node;
Inserting a new node Node* List: : Insert. Node(int index, double x) { if (index < 0) return NULL; int curr. Index = 1; Node* curr. Node = head; while (curr. Node && index > curr. Index) { curr. Node = curr. Node->next; curr. Index++; } if (index > 0 && curr. Node == NULL) return NULL; Node* new. Node = new. Node->data = if (index == 0) { new. Node->next head } else { new. Node->next curr. Node->next } return new. Node; } new x; Node; = = head; new. Node; = = curr. Node->next; new. Node; Create a new node
Inserting a new node Node* List: : Insert. Node(int index, double x) { if (index < 0) return NULL; int curr. Index = 1; Node* curr. Node = head; while (curr. Node && index > curr. Index) { curr. Node = curr. Node->next; curr. Index++; } if (index > 0 && curr. Node == NULL) return NULL; Node* new. Node = new. Node->data = if (index == 0) { new. Node->next head } else { new. Node->next curr. Node->next } return new. Node; } new x; Node; = = head; new. Node; = = curr. Node->next; new. Node; Insert as first element head new. Node
Inserting a new node Node* List: : Insert. Node(int index, double x) { if (index < 0) return NULL; int curr. Index = 1; Node* curr. Node = head; while (curr. Node && index > curr. Index) { curr. Node = curr. Node->next; curr. Index++; } if (index > 0 && curr. Node == NULL) return NULL; Node* new. Node = new. Node->data = if (index == 0) { new. Node->next head } else { new. Node->next curr. Node->next } return new. Node; new x; Node; = = head; new. Node; Insert after curr. Node = = curr. Node->next; curr. Node new. Node; } new. Node
Finding a node • int Find. Node(double x) – Search for a node with the value equal to x in the list. – If such a node is found, return its position. Otherwise, return 0. int List: : Find. Node(double x) { Node* curr. Node = head; int curr. Index = 1; while (curr. Node && curr. Node->data != x) { curr. Node = curr. Node->next; curr. Index++; } if (curr. Node) return curr. Index; return 0; }
Deleting a node • int Delete. Node(double x) – Delete a node with the value equal to x from the list. – If such a node is found, return its position. Otherwise, return 0. • Steps – Find the desirable node (similar to Find. Node) – Release the memory occupied by the found node – Set the pointer of the predecessor of the found node to the successor of the found node • Like Insert. Node, there are two special cases – Delete first node – Delete the node in middle or at the end of the list
The Scenario head 6 4 17 • Begin with an existing linked list – Could be empty or not – Could be ordered or not 42
The Scenario head 6 4 17 • Begin with an existing linked list – Could be empty or not – Could be ordered or not 42
The Scenario head 4 17 42 • Begin with an existing linked list – Could be empty or not – Could be ordered or not
The Scenario head 4 17 42 • Begin with an existing linked list – Could be empty or not – Could be ordered or not
Deleting a node int List: : Delete. Node(double x) { Try to find the node with its Node* prev. Node = NULL; value equal to x Node* curr. Node = head; int curr. Index = 1; while (curr. Node && curr. Node->data != x) { prev. Node = curr. Node; curr. Node = curr. Node->next; curr. Index++; } if (curr. Node) { if (prev. Node) { prev. Node->next = curr. Node->next; delete curr. Node; } else { head = curr. Node->next; delete curr. Node; } return curr. Index; } return 0; }
Deleting a node int List: : Delete. Node(double x) { Node* prev. Node = NULL; Node* curr. Node = head; int curr. Index = 1; while (curr. Node && curr. Node->data != x) { prev. Node = curr. Node; curr. Node = curr. Node->next; curr. Index++; prev. Node curr. Node } if (curr. Node) { if (prev. Node) { prev. Node->next = curr. Node->next; delete curr. Node; } else { head = curr. Node->next; delete curr. Node; } return curr. Index; } return 0; }
Deleting a node int List: : Delete. Node(double x) { Node* prev. Node = NULL; Node* curr. Node = head; int curr. Index = 1; while (curr. Node && curr. Node->data != x) { prev. Node = curr. Node; curr. Node = curr. Node->next; curr. Index++; } if (curr. Node) { if (prev. Node) { prev. Node->next = curr. Node->next; delete curr. Node; } else { head = curr. Node->next; delete curr. Node; } return curr. Index; } head curr. Node return 0; }
Printing all the elements • void Display. List(void) – Print the data of all the elements – Print the number of the nodes in the list void List: : Display. List() { int num = 0; Node* curr. Node = head; while (curr. Node != NULL){ cout << curr. Node->data << endl; curr. Node = curr. Node->next; num++; } cout << "Number of nodes in the list: " << num << endl; }
Destroying the list • ~List(void) – Use the destructor to release all the memory used by the list. – Step through the list and delete each node one by one. List: : ~List(void) { Node* curr. Node = head, *next. Node = NULL; while (curr. Node != NULL) { next. Node = curr. Node->next; // destroy the current node delete curr. Node; curr. Node = next. Node; } }
Using List int main(void) { List list; list. Insert. Node(0, 7. 0); list. Insert. Node(1, 5. 0); list. Insert. Node(-1, 5. 0); list. Insert. Node(0, 6. 0); list. Insert. Node(8, 4. 0); // print all the elements list. Display. List(); if(list. Find. Node(5. 0) > 0) else if(list. Find. Node(4. 5) > 0) else list. Delete. Node(7. 0); list. Display. List(); return 0; } // // // 6 result 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 successful unsuccessful cout << << "5. 0 "4. 5 found" << endl; not found" << endl;
Linked List Efficiency • Insertion and deletion at the beginning of the list are very fast, O(1). • Finding, deleting or inserting in the list requires searching through half the items in the list on an average, requiring O(n) comparisons. • Although arrays require same number of comparisons, the advantage lies in the fact that no items need to be moved after insertion or deletion. • As opposed to fixed size of arrays, linked lists use exactly as much memory as is needed and can expand.
More on Linked Lists
Summary list § is a sequential container optimized for insertion and erasure at arbitrary points in the sequence Linked List § Each data item is embedded in a link & Each Link object contains a reference to the next link in the list of items.
Circular Linked List In other applications a circular linked list is used; instead of the last node containing a null pointer, it contains a pointer to the first node in the list. For such lists, one can use a single pointer to the last node in the list, because then one has direct access to it and "almost-direct" access to the first node. last 9 17 22 26 34 §Each node in a circular linked list has a predecessor (and a successor), provided that the list is nonempty. §insertion and deletion do not require special consideration of the first node. §This is a good implementation for a linked queue or for any problem in which "circular" processing is required
Node* prev. Node = NULL; Node* curr. Node = head; int curr. Index = 1; while (curr. Node && curr. Node->data != x) { prev. Node = curr. Node; curr. Node = curr. Node->next; curr. Index++; }
Inserting a new node Node* List: : Insert. Node(int index, double x) { if (index < 0) return NULL; Try to locate index’th node. If it doesn’t exist, return NULL. int curr. Index = 1; Node* curr. Node = head; while (curr. Node && index > curr. Index) { prevnode=currnode curr. Node = curr. Node->next; curr. Index++; } if (index > 0 && curr. Node == NULL) return NULL; Node* new. Node = new. Node->data = if (index == 0) { new. Node->next head } else { new. Node->next curr. Node->next } return new. Node; } new x; Node; = = head; new. Node; = = curr. Node->next; new. Node;
Circularly Linked Lists For example, item can be inserted as follows: newptr = new Node(item, 0); if (Index == 0) // list is empty { newptr->next = newptr; current. Index = newptr; } else // nonempty list { newptr->next = predptr->next; predptr->next = newptr; } Note that a one-element circularly linked list points to itself.
Circularly Linked Lists Traversal must be modified: avoid an infinite loop by looking for the end of list as signalled by a null pointer. Like other methods, deletion must also be slightly modified. Deleting the last node is signalled when the node deleted points to itself. if (Index == 0) // list is empty // Signal that the list is empty else { ptr = predptr->next; // hold node for deletion if (ptr == predptr) // one-node list Index = 0; else // list with 2 or more nodes predptr->next = ptr->next; delete ptr; }
Summary In Circular linked lists – The last node points to the first node of the list A B C Head – 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. )
Double Linked Lists
Doubly Linked Lists All of these lists, however, are uni-directional; we can only move from one node to its successor. In many applications, bidirectional movement is necessary. In this case, each node has two pointers — one to its successor (null if there is none) and one to its predecessor (null if there is none. ) Such a list is commonly called a doubly-linked (or symmetrically-linked) list. L prev last 9 first my. Size 5 next 17 22 26 34
Con’t…. • Similar to an ordinary list with the addition that a link to the last item is maintained along with that to the first. • The reference to the last link permits to insert a new link directly at the end of the list as well as at the beginning. • This could not be done in the ordinary linked list without traversing the whole list. • This technique is useful in implementing the Queue where insertions are made at end and deletions from the front.
Insertion • We visualize operation insert. After(p, X), which returns position q p A B C p A q B C X p A q B Linked Lists X C 41
Insertion Algorithm insert. After(p, e): Create a new node v v. set. Element(e) v. set. Prev(p){link v to its predecessor} v. set. Next(p. get. Next()) {link v to its successor} (p. get. Next()). set. Prev(v) {link p’s old successor to v} p. set. Next(v) return v {link p to its new successor, v} {the position for the element e} Linked Lists 42
Deletion • We visualize remove(p), where p == last() A B C p D A B Linked Lists C 43
Deletion Algorithm remove(p): t = p. element {a temporary variable to hold the return value} (p. get. Prev()). set. Next(p. get. Next()) {linking out p} (p. get. Next()). set. Prev(p. get. Prev()) p. set. Prev(null) {invalidating the position p} p. set. Next(null) return t Linked Lists 44
Worst-cast running time • In a doubly linked list + insertion at head or tail is in O(1) + deletion at either end is on O(1) -- element access is still in O(n) Linked Lists 45
Summary – Each node points to not only successor but the predecessor – There are two NULL: at the first and last nodes in the list – Advantage: given a node, it is easy to visit its predecessor. Convenient to traverse lists backwards
Summary • A doubly linked list is often more convenient! • Nodes store: – element – link to the previous node – link to the next node prev next elem node • Special trailer and header nodes/positions header trailer elements Linked Lists 47
Con’t…. doubly-linked list => bidirectional movement!! Each node has two pointers — one to its successor (null if there is none) and one to its predecessor (null if there is none. ) Doubly linked lists give one more flexibility (can move in either direction) BUT at significant cost : DOUBLE the overhead for links More complex code
The End !!!
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