Can linked list contains duplicate values?

In this program, we need to remove the duplicate nodes from the given singly linked list.

Original List:

List after removing duplicate nodes:

In the above list, node 2 is repeated thrice, and node 1 is repeated twice. Node 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 current to another node temp. Connect temp's next node with index's next node. Delete index which was pointing to duplicate node. This process will continue until all duplicates are removed.

Algorithm

1. Create a class Node which has two attributes: data and next. Next is a pointer to the next node in the list.
2. Create another class RemoveDuplicate which has two attributes: head and tail.
3. addNode[] will add a new node to the list:
   1. Create a new node.
   2. It first checks, whether the head is equal to null which means the list is empty.
   3. If the list is empty, both head and tail will point to a newly added node.
   4. If the list is not empty, the new node will be added to end of the list such that tail's next will point to a newly added node. This new node will become the new tail of the list.
4. removeDuplicate[] will remove duplicate nodes from the list.
   1. Define a new node current which will initially point to head.
   2. Node temp will point to current and index will always point to node next to current.
   3. Loop through the list till current points to null.
   4. Check whether current?s data is equal to index's data that means index is duplicate of current.
   5. Since index points to duplicate node so skip it by making node next to temp to will point to node next to index, i.e. temp.next = index.next.
5. display[] will display the nodes present in the list:
   1. Define a node current which will initially point to the head of the list.
   2. Traverse through the list till current points to null.
   3. Display each node by making current to point to node next to it in each iteration.

Solution

Python

#Represent a node of the singly linked list class Node: def __init__[self,data]: self.data = data; self.next = None; class RemoveDuplicate: #Represent the head and tail of the singly linked list def __init__[self]: self.head = None; self.tail = None; #addNode[] will add a new node to the list def addNode[self, data]: #Create a new node newNode = Node[data]; #Checks if the list is empty if[self.head == None]: #If list is empty, both head and tail will point to new node self.head = newNode; self.tail = newNode; else: #newNode will be added after tail such that tail's next will point to newNode self.tail.next = newNode; #newNode will become new tail of the list self.tail = newNode; #removeDuplicate[] will remove duplicate nodes from the list def removeDuplicate[self]: #Node current will point to head current = self.head; index = None; temp = None; if[self.head == None]: return; else: while[current != None]: #Node temp will point to previous node to index. temp = current; #Index will point to node next to current index = current.next; while[index != None]: #If current node's data is equal to index node's data if[current.data == index.data]: #Here, index node is pointing to the node which is duplicate of current node #Skips the duplicate node by pointing to next node temp.next = index.next; else: #Temp will point to previous node of index. temp = index; index = index.next; current = current.next; #display[] will display all the nodes present in 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; sList = RemoveDuplicate[]; #Adds data to the list sList.addNode[1]; sList.addNode[2]; sList.addNode[3]; sList.addNode[2]; sList.addNode[2]; sList.addNode[4]; sList.addNode[1]; print["Originals list: "]; sList.display[]; #Removes duplicate nodes sList.removeDuplicate[]; print["List after removing duplicates: "]; sList.display[];

Output:

Originals list: 1 2 3 2 2 4 1 List after removing duplicates: 1 2 3 4

C

#include //Represent a node of the singly linked list struct node{ int data; struct node *next; }; //Represent the head and tail of the singly linked list struct node *head, *tail = NULL; //addNode[] will add a new node to the list void addNode[int data] { //Create a new node struct node *newNode = [struct node*]malloc[sizeof[struct node]]; newNode->data = data; newNode->next = NULL; //Checks if the list is empty if[head == NULL] { //If list is empty, both head and tail will point to new node head = newNode; tail = newNode; } else { //newNode will be added after tail such that tail's next will point to newNode tail->next = newNode; //newNode will become new tail of the list tail = newNode; } } //removeDuplicate[] will remove duplicate nodes from the list void removeDuplicate[] { //Node current will point to head struct node *current = head, *index = NULL, *temp = NULL; if[head == NULL] { return; } else { while[current != NULL]{ //Node temp will point to previous node to index. temp = current; //Index will point to node next to current index = current->next; while[index != NULL] { //If current node's data is equal to index node's data if[current->data == index->data] { //Here, index node is pointing to the node which is duplicate of current node //Skips the duplicate node by pointing to next node temp->next = index->next; } else { //Temp will point to previous node of index. temp = index; } index = index->next; } current = current->next; } } } //display[] will display all the nodes present in the list void display[] { //Node current will point to head struct node *current = head; if[head == NULL] { printf["List is empty \n"]; return; } while[current != NULL] { //Prints each node by incrementing pointer printf["%d ", current->data]; current = current->next; } printf["\n"]; } int main[] { //Adds data to the list addNode[1]; addNode[2]; addNode[3]; addNode[2]; addNode[2]; addNode[4]; addNode[1]; printf["Originals list: \n"]; display[]; //Removes duplicate nodes removeDuplicate[]; printf["List after removing duplicates: \n"]; display[]; return 0; }

Output:

Originals list: 1 2 3 2 2 4 1 List after removing duplicates: 1 2 3 4

JAVA

public class RemoveDuplicate { //Represent a node of the singly linked list class Node{ int data; Node next; public Node[int data] { this.data = data; this.next = null; } } //Represent the head and tail of the singly linked list public Node head = null; public Node tail = null; //addNode[] will add a new node to the list public void addNode[int data] { //Create a new node Node newNode = new Node[data]; //Checks if the list is empty if[head == null] { //If list is empty, both head and tail will point to new node head = newNode; tail = newNode; } else { //newNode will be added after tail such that tail's next will point to newNode tail.next = newNode; //newNode will become new tail of the list tail = newNode; } } //removeDuplicate[] will remove duplicate nodes from the list public void removeDuplicate[] { //Node current will point to head Node current = head, index = null, temp = null; if[head == null] { return; } else { while[current != null]{ //Node temp will point to previous node to index. temp = current; //Index will point to node next to current index = current.next; while[index != null] { //If current node's data is equal to index node's data if[current.data == index.data] { //Here, index node is pointing to the node which is duplicate of current node //Skips the duplicate node by pointing to next node temp.next = index.next; } else { //Temp will point to previous node of index. temp = index; } index = index.next; } current = current.next; } } } //display[] will display all the nodes present in the list public void display[] { //Node current will point to head Node current = head; if[head == null] { System.out.println["List is empty"]; return; } while[current != null] { //Prints each node by incrementing pointer System.out.print[current.data + " "]; current = current.next; } System.out.println[]; } public static void main[String[] args] { RemoveDuplicate sList = new RemoveDuplicate[]; //Adds data to the list sList.addNode[1]; sList.addNode[2]; sList.addNode[3]; sList.addNode[2]; sList.addNode[2]; sList.addNode[4]; sList.addNode[1]; System.out.println["Originals list: "]; sList.display[]; //Removes duplicate nodes sList.removeDuplicate[]; System.out.println["List after removing duplicates: "]; sList.display[]; } }

Output:

Originals list: 1 2 3 2 2 4 1 List after removing duplicates: 1 2 3 4

C#

using System; public class CreateList { //Represent a node of the singly linked list public class Node{ public T data; public Node next; public Node[T value] { data = value; next = null; } } public class RemoveDuplicate{ //Represent the head and tail of the singly linked list public Node head = null; public Node tail = null; //addNode[] will add a new node to the list public void addNode[T data] { //Create a new node Node newNode = new Node[data]; //Checks if the list is empty if[head == null] { //If list is empty, both head and tail will point to new node head = newNode; tail = newNode; } else { //newNode will be added after tail such that tail's next will point to newNode tail.next = newNode; //newNode will become new tail of the list tail = newNode; } } //removeDuplicate[] will remove duplicate nodes from the list public void removeDuplicate[] { //Node current will point to head Node current = head, index = null, temp = null; if[head == null] { return; } else { while[current != null]{ //Node temp will point to previous node to index. temp = current; //Index will point to node next to current index = current.next; while[index != null] { //If current node's data is equal to index node's data if[current.data.Equals[index.data]] { //Here, index node is pointing to the node which is duplicate of current node //Skips the duplicate node by pointing to next node temp.next = index.next; } else { //Temp will point to previous node of index. temp = index; } index = index.next; } current = current.next; } } } //display[] will display all the nodes present in the list public void display[] { //Node current will point to head Node current = head; if[head == null] { Console.WriteLine["List is empty"]; return; } while[current != null] { //Prints each node by incrementing pointer Console.Write[current.data + " "]; current = current.next; } Console.WriteLine[]; } } public static void Main[] { RemoveDuplicate sList = new RemoveDuplicate[]; //Adds data to the list sList.addNode[1]; sList.addNode[2]; sList.addNode[3]; sList.addNode[2]; sList.addNode[2]; sList.addNode[4]; sList.addNode[1]; Console.WriteLine["Originals list: "]; sList.display[]; //Removes duplicate nodes sList.removeDuplicate[]; Console.WriteLine["List after removing duplicates: "]; sList.display[]; } }

Output:

Originals list: 1 2 3 2 2 4 1 List after removing duplicates: 1 2 3 4

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