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Linked Lists Geletaw S.

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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

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Array vs Linked List node node Array node Linked List

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**What’s wrong with Array ?**

Disadvantages of arrays as storage data structures slow searching in unordered array slow insertion in ordered array Fixed size

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**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.

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**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

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**Linked Lists A linked list is a series of connected nodes**

Head B C 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 A data pointer node

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**A Simple Linked List Class**

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

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**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 };

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**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; } // 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; };

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**Inserting a new node Node* InsertNode(int index, double x) Steps**

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 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 newNode

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**Inserting a new node Possible cases of InsertNode**

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)

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Inserting a new node Try to locate index’th node. If it doesn’t exist, return NULL. 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 = new Node; newNode->data = x; if (index == 0) { newNode->next = head; head = newNode; else { newNode->next = currNode->next; currNode->next = newNode; return newNode;

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**Inserting a new node Create a new 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 = new Node; 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

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**Inserting a new node Insert as first element**

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 = new Node; 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

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**Inserting a new node Insert after currNode**

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 = new Node; newNode->data = x; if (index == 0) { newNode->next = head; head = newNode; else { newNode->next = currNode->next; currNode->next = newNode; return newNode; Insert after currNode currNode newNode

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**Finding a node int FindNode(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::FindNode(double x) { Node* currNode = head; int currIndex = 1; while (currNode && currNode->data != x) { currNode = currNode->next; currIndex++; } if (currNode) return currIndex; return 0;

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**Deleting a node Steps Like InsertNode, there are two special cases**

int DeleteNode(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 FindNode) 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 InsertNode, there are two special cases Delete first node Delete the node in middle or at the end of the list

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**The Scenario Begin with an existing linked list Could be empty or not**

head 6 42 4 17 Begin with an existing linked list Could be empty or not Could be ordered or not

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**The Scenario Begin with an existing linked list Could be empty or not**

head 6 42 4 17 Begin with an existing linked list Could be empty or not Could be ordered or not

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**The Scenario Begin with an existing linked list Could be empty or not**

head 42 4 17 Begin with an existing linked list Could be empty or not Could be ordered or not

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**The Scenario Begin with an existing linked list Could be empty or not**

head 42 4 17 Begin with an existing linked list Could be empty or not Could be ordered or not

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**Deleting a node Try to find the node with its value equal to x**

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; return currIndex; return 0; Try to find the node with its value equal to x

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**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; return currIndex; return 0; prevNode currNode

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**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; return currIndex; return 0; head currNode

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**Printing all the elements**

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

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**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* currNode = head, *nextNode = NULL; while (currNode != NULL) { nextNode = currNode->next; // destroy the current node delete currNode; currNode = nextNode; }

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**Using List result int main(void) { List list;**

6 7 5 Number of nodes in the list: 3 5.0 found 4.5 not found Number of nodes in the list: 2 result 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; else cout << "5.0 not found" << endl; if(list.FindNode(4.5) > 0) cout << "4.5 found" << endl; else cout << "4.5 not found" << endl; list.DeleteNode(7.0); return 0; }

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**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.

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More on Linked Lists

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**Summary list Linked 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.

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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

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Node* prevNode = NULL; Node* currNode = head; int currIndex = 1; while (currNode && currNode->data != x) { prevNode = currNode; currNode = currNode->next; currIndex++; }

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

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**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; currentIndex = newptr; } else // nonempty list newptr->next = predptr->next; predptr->next = newptr; Note that a one-element circularly linked list points to itself.

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**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; }

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**Summary In Circular linked lists**

The last node points to the first node of the list 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 B C Head

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Double Linked Lists

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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. last prev L first mySize 5 9 17 22 26 34 next

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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.

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Insertion We visualize operation insertAfter(p, X), which returns position q p A B C p q A B C X p q A B X C Linked Lists

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**Insertion Algorithm Algorithm insertAfter(p,e): Create a new node v**

v.setElement(e) v.setPrev(p) {link v to its predecessor} v.setNext(p.getNext()) {link v to its successor} (p.getNext()).setPrev(v) {link p’s old successor to v} p.setNext(v) {link p to its new successor, v} return v {the position for the element e} Linked Lists

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**Deletion We visualize remove(p), where p == last() A B C D p A B C p D**

Linked Lists

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**Deletion Algorithm Algorithm remove(p):**

t = p.element {a temporary variable to hold the return value} (p.getPrev()).setNext(p.getNext()) {linking out p} (p.getNext()).setPrev(p.getPrev()) p.setPrev(null) {invalidating the position p} p.setNext(null) return t Linked Lists

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**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

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**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

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**Summary A doubly linked list is often more convenient! Nodes store:**

element link to the previous node link to the next node Special trailer and header nodes prev next elem node trailer header nodes/positions elements Linked Lists

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**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

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The End !!!

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