Python program to remove duplicate elements from a Doubly Linked List

In this program, we will create a doubly linked list and remove the duplicate, if present, by traversing through the list.

List after removing duplicates:

In the above list, node2 is repeated thrice, and node 3 is repeated twice. Current will point to head, and index will point to node next to current. Start traversing the list till a duplicate is found that is when current's data is equal to index's data. In the above example, the first duplicate will be found at position 4. Assign index to another node temp. Connect index's previous node with index's next node. Delete temp which was pointing to duplicate node. This process will continue till all duplicates are removed.

ALGORITHM:

Define a Node class which represents a node in the list. It will have three properties: data, previous which will point to the previous node and next which will point to the next node.
Define another class for creating a doubly linked list, and it has two nodes: head and tail. Initially, head and tail will point to null.
addNode() will add node to the list:

It first checks whether the head is null, then it will insert the node as the head.
Both head and tail will point to a newly added node.
Head's previous pointer will point to null and tail's next pointer will point to null.
If the head is not null, the new node will be inserted at the end of the list such that new node's previous pointer will point to tail.
The new node will become the new tail. Tail's next pointer will point to null.

a. removeDuplicateNode() will remove duplicate nodes from the list.

Define a new node current which will initially point to head.
Node index will always point to node next to current.
Loop through the list until current points to null.
Check whether current's data is equal to index's data that means index is duplicate of current.
Node temp will point to index to store duplicate node.
Since, temp is pointing to index, which is a duplicate node, so set temp to null.

a. display() will show all the nodes present in the list.

Define a new node 'current' that will point to the head.
Print current.data till current points to null.
Current will point to the next node in the list in each iteration.

#Represent a node of doubly linked list class Node: def __init__(self,data): self.data = data; self.previous = None; self.next = None; class RemoveDuplicate: #Represent the head and tail of the doubly linked list def __init__(self): self.head = None; self.tail = None; #addNode() will add a node to the list def addNode(self, data): #Create a new node newNode = Node(data); #If list is empty if(self.head == None): #Both head and tail will point to newNode self.head = self.tail = newNode; #head's previous will point to None self.head.previous = None; #tail's next will point to None, as it is the last node of the list self.tail.next = None; else: #newNode will be added after tail such that tail's next will point to newNode self.tail.next = newNode; #newNode's previous will point to tail newNode.previous = self.tail; #newNode will become new tail self.tail = newNode; #As it is last node, tail's next will point to None self.tail.next = None; #removeDuplicateNode() will remove duplicate nodes from the list def removeDuplicateNode(self): #Checks whether list is empty if(self.head == None): return; else: #Initially, current will point to head node current = self.head; while(current != None): #index will point to node next to current index = current.next while(index != None): if(current.data == index.data): #Store the duplicate node in temp temp = index; #index's previous node will point to node next to index thus, removes the duplicate node index.previous.next = index.next; if(index.next != None): index.next.previous = index.previous; #Delete duplicate node by making temp to None temp = None; index = index.next; current = current.next; #display() will print out the nodes of the list def display(self): #Node current will point to head current = self.head; if(self.head == None): print("List is empty"); return; while(current != None): #Prints each node by incrementing pointer. print(current.data), current = current.next; print(); dList = RemoveDuplicate(); #Add nodes to the list dList.addNode(1); dList.addNode(2); dList.addNode(3); dList.addNode(2); dList.addNode(2); dList.addNode(4); dList.addNode(5); dList.addNode(3); print("Originals list: "); dList.display(); #Removes duplicate nodes dList.removeDuplicateNode(); print("List after removing duplicates: "); dList.display();

Python program to remove duplicate elements from a Doubly Linked List

ALGORITHM:

All Answers

PROGRAM:

Python Doubly Linked List Programs

This question belongs to these collections

Similar questions

need a help?

find thousands of online teachers now