Remove Duplicates from an Unsorted Linked List
Given an unsorted Linked List, the task is to remove duplicates from the list.
Examples:
Input: linked_list = 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Output: 12 -> 11 -> 21 -> 41 -> 43
Explanation: Second occurrence of 12 and 21 are removed.
Input: linked_list = 12 -> 11 -> 12 -> 21 -> 41 -> 43 -> 21
Output: 12 -> 11 -> 21 -> 41 -> 43
Naive Approach to Remove Duplicates from an Unsorted Linked List:
The most simple approach to solve this, is to check each node for duplicate in the Linked List one by one.
Below is the Implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node* next;
};
struct Node* newNode( int data)
{
Node* temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
}
void removeDuplicates( struct Node* start)
{
struct Node *ptr1, *ptr2, *dup;
ptr1 = start;
while (ptr1 != NULL && ptr1->next != NULL) {
ptr2 = ptr1;
while (ptr2->next != NULL) {
if (ptr1->data == ptr2->next->data) {
dup = ptr2->next;
ptr2->next = ptr2->next->next;
delete (dup);
}
else
ptr2 = ptr2->next;
}
ptr1 = ptr1->next;
}
}
void printList( struct Node* node)
{
while (node != NULL) {
printf ( "%d " , node->data);
node = node->next;
}
}
int main()
{
struct Node* start = newNode(10);
start->next = newNode(12);
start->next->next = newNode(11);
start->next->next->next = newNode(11);
start->next->next->next->next = newNode(12);
start->next->next->next->next->next = newNode(11);
start->next->next->next->next->next->next = newNode(10);
printf ( "Linked list before removing duplicates " );
printList(start);
removeDuplicates(start);
printf ( "\nLinked list after removing duplicates " );
printList(start);
return 0;
}
|
C
#include <stdio.h>
#include <stdlib.h>
typedef struct Node {
int data;
struct Node* next;
} Node;
Node* newNode( int data)
{
Node* temp = (Node*) malloc ( sizeof (Node));
temp->data = data;
temp->next = NULL;
return temp;
}
void removeDuplicates(Node* start)
{
Node *ptr1, *ptr2, *dup;
ptr1 = start;
while (ptr1 != NULL && ptr1->next != NULL) {
ptr2 = ptr1;
while (ptr2->next != NULL) {
if (ptr1->data == ptr2->next->data) {
dup = ptr2->next;
ptr2->next = ptr2->next->next;
free (dup);
}
else
ptr2 = ptr2->next;
}
ptr1 = ptr1->next;
}
}
void printList( struct Node* node)
{
while (node != NULL) {
printf ( "%d " , node->data);
node = node->next;
}
}
int main()
{
struct Node* start = newNode(10);
start->next = newNode(12);
start->next->next = newNode(11);
start->next->next->next = newNode(11);
start->next->next->next->next = newNode(12);
start->next->next->next->next->next = newNode(11);
start->next->next->next->next->next->next = newNode(10);
printf ( "Linked list before removing duplicates " );
printList(start);
removeDuplicates(start);
printf ( "\nLinked list after removing duplicates " );
printList(start);
return 0;
}
|
Java
class LinkedList {
static Node head;
static class Node {
int data;
Node next;
Node( int d)
{
data = d;
next = null ;
}
}
void remove_duplicates()
{
Node ptr1 = null , ptr2 = null , dup = null ;
ptr1 = head;
while (ptr1 != null && ptr1.next != null ) {
ptr2 = ptr1;
while (ptr2.next != null ) {
if (ptr1.data == ptr2.next.data) {
ptr2.next = ptr2.next.next;
System.gc();
}
else {
ptr2 = ptr2.next;
}
}
ptr1 = ptr1.next;
}
}
void printList(Node node)
{
while (node != null ) {
System.out.print(node.data + " " );
node = node.next;
}
}
public static void main(String[] args)
{
LinkedList list = new LinkedList();
list.head = new Node( 10 );
list.head.next = new Node( 12 );
list.head.next.next = new Node( 11 );
list.head.next.next.next = new Node( 11 );
list.head.next.next.next.next = new Node( 12 );
list.head.next.next.next.next.next = new Node( 11 );
list.head.next.next.next.next.next.next
= new Node( 10 );
System.out.println(
"Linked List before removing duplicates : " );
list.printList(head);
list.remove_duplicates();
System.out.println( "\n" );
System.out.println(
"Linked List after removing duplicates : " );
list.printList(head);
}
}
|
Python3
class Node():
def __init__( self , data):
self .data = data
self . next = None
class LinkedList():
def __init__( self ):
self .head = None
def remove_duplicates( self ):
ptr1 = None
ptr2 = None
dup = None
ptr1 = self .head
while (ptr1 ! = None and ptr1. next ! = None ):
ptr2 = ptr1
while (ptr2. next ! = None ):
if (ptr1.data = = ptr2. next .data):
dup = ptr2. next
ptr2. next = ptr2. next . next
else :
ptr2 = ptr2. next
ptr1 = ptr1. next
def printList( self ):
temp = self .head
while (temp ! = None ):
print (temp.data, end = " " )
temp = temp. next
print ()
list = LinkedList()
list .head = Node( 10 )
list .head. next = Node( 12 )
list .head. next . next = Node( 11 )
list .head. next . next . next = Node( 11 )
list .head. next . next . next . next = Node( 12 )
list .head. next . next . next . next . next = Node( 11 )
list .head. next . next . next . next . next . next = Node( 10 )
print ( "Linked List before removing duplicates :" )
list .printList()
list .remove_duplicates()
print ()
print ( "Linked List after removing duplicates :" )
list .printList()
|
C#
using System;
class List_ {
Node head;
class Node {
public int data;
public Node next;
public
Node( int d)
{
data = d;
next = null ;
}
}
void remove_duplicates()
{
Node ptr1 = null , ptr2 = null ;
ptr1 = head;
while (ptr1 != null && ptr1.next != null ) {
ptr2 = ptr1;
while (ptr2.next != null ) {
if (ptr1.data == ptr2.next.data) {
ptr2.next = ptr2.next.next;
}
else
{
ptr2 = ptr2.next;
}
}
ptr1 = ptr1.next;
}
}
void printList(Node node)
{
while (node != null ) {
Console.Write(node.data + " " );
node = node.next;
}
}
public static void Main(String[] args)
{
List_ list = new List_();
list.head = new Node(10);
list.head.next = new Node(12);
list.head.next.next = new Node(11);
list.head.next.next.next = new Node(11);
list.head.next.next.next.next = new Node(12);
list.head.next.next.next.next.next = new Node(11);
list.head.next.next.next.next.next.next
= new Node(10);
Console.WriteLine(
"Linked List_ before removing duplicates: " );
list.printList(list.head);
list.remove_duplicates();
Console.WriteLine( "" );
Console.WriteLine(
"Linked List_ after removing duplicates: " );
list.printList(list.head);
}
}
|
Javascript
<script>
var head;
class Node {
constructor(val) {
this .data = val;
this .next = null ;
}
}
function remove_duplicates() {
var ptr1 = null , ptr2 = null , dup = null ;
ptr1 = head;
while (ptr1 != null && ptr1.next != null ) {
ptr2 = ptr1;
while (ptr2.next != null ) {
if (ptr1.data == ptr2.next.data) {
dup = ptr2.next;
ptr2.next = ptr2.next.next;
} else {
ptr2 = ptr2.next;
}
}
ptr1 = ptr1.next;
}
}
function printList( node) {
while (node != null ) {
console.log(node.data + " " );
node = node.next;
}
}
head = new Node(10);
head.next = new Node(12);
head.next.next = new Node(11);
head.next.next.next = new Node(11);
head.next.next.next.next = new Node(12);
head.next.next.next.next.next = new Node(11);
head.next.next.next.next.next.next = new Node(10);
document.write( "Linked List before removing duplicates : <br/> " );
printList(head);
remove_duplicates();
document.write( "<br/>" );
document.write( "Linked List after removing duplicates : <br/> " );
printList(head);
</script>
|
Output
Linked list before removing duplicates 10 12 11 11 12 11 10
Linked list after removing duplicates 10 12 11
Time Complexity: O(N2)
Auxiliary Space: O(1)
Remove duplicates from an Unsorted Linked List using Hashing:
The idea for this approach is based on the following observations:
- Traverse the link list from head to end.
- For every newly encountered element, check whether if it is in the hash table:
- if yes, remove it
- otherwise put it in the hash table.
- At the end, the Hash table will contain only the unique elements.
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
struct Node {
int data;
struct Node* next;
};
struct Node* newNode( int data)
{
Node* temp = new Node;
temp->data = data;
temp->next = NULL;
return temp;
}
void removeDuplicates( struct Node* start)
{
unordered_set< int > seen;
struct Node* curr = start;
struct Node* prev = NULL;
while (curr != NULL) {
if (seen.find(curr->data) != seen.end()) {
prev->next = curr->next;
delete (curr);
}
else {
seen.insert(curr->data);
prev = curr;
}
curr = prev->next;
}
}
void printList( struct Node* node)
{
while (node != NULL) {
printf ( "%d " , node->data);
node = node->next;
}
}
int main()
{
struct Node* start = newNode(10);
start->next = newNode(12);
start->next->next = newNode(11);
start->next->next->next = newNode(11);
start->next->next->next->next = newNode(12);
start->next->next->next->next->next = newNode(11);
start->next->next->next->next->next->next = newNode(10);
printf ( "Linked list before removing duplicates : \n" );
printList(start);
removeDuplicates(start);
printf ( "\nLinked list after removing duplicates : \n" );
printList(start);
return 0;
}
|
Java
import java.util.HashSet;
public class removeDuplicates {
static class node {
int val;
node next;
public node( int val) { this .val = val; }
}
static void removeDuplicate(node head)
{
HashSet<Integer> hs = new HashSet<>();
node current = head;
node prev = null ;
while (current != null ) {
int curval = current.val;
if (hs.contains(curval)) {
prev.next = current.next;
}
else {
hs.add(curval);
prev = current;
}
current = current.next;
}
}
static void printList(node head)
{
while (head != null ) {
System.out.print(head.val + " " );
head = head.next;
}
}
public static void main(String[] args)
{
node start = new node( 10 );
start.next = new node( 12 );
start.next.next = new node( 11 );
start.next.next.next = new node( 11 );
start.next.next.next.next = new node( 12 );
start.next.next.next.next.next = new node( 11 );
start.next.next.next.next.next.next = new node( 10 );
System.out.println(
"Linked list before removing duplicates :" );
printList(start);
removeDuplicate(start);
System.out.println(
"\nLinked list after removing duplicates :" );
printList(start);
}
}
|
Python3
class Node:
def __init__( self , data):
self .data = data
self . next = None
class LinkedList:
def __init__( self ):
self .head = None
def printlist( self ):
temp = self .head
while (temp):
print (temp.data, end = " " )
temp = temp. next
def removeDuplicates( self , head):
if self .head is None or self .head. next is None :
return head
hash = set ()
current = head
hash .add( self .head.data)
while current. next is not None :
if current. next .data in hash :
current. next = current. next . next
else :
hash .add(current. next .data)
current = current. next
return head
if __name__ = = "__main__" :
llist = LinkedList()
llist.head = Node( 10 )
second = Node( 12 )
third = Node( 11 )
fourth = Node( 11 )
fifth = Node( 12 )
sixth = Node( 11 )
seventh = Node( 10 )
llist.head. next = second
second. next = third
third. next = fourth
fourth. next = fifth
fifth. next = sixth
sixth. next = seventh
print ( "Linked List before removing Duplicates." )
llist.printlist()
llist.removeDuplicates(llist.head)
print ( "\nLinked List after removing duplicates." )
llist.printlist()
|
C#
using System;
using System.Collections.Generic;
class removeDuplicates {
class node {
public int val;
public node next;
public node( int val) { this .val = val; }
}
static void removeDuplicate(node head)
{
HashSet< int > hs = new HashSet< int >();
node current = head;
node prev = null ;
while (current != null ) {
int curval = current.val;
if (hs.Contains(curval)) {
prev.next = current.next;
}
else {
hs.Add(curval);
prev = current;
}
current = current.next;
}
}
static void printList(node head)
{
while (head != null ) {
Console.Write(head.val + " " );
head = head.next;
}
}
public static void Main(String[] args)
{
node start = new node(10);
start.next = new node(12);
start.next.next = new node(11);
start.next.next.next = new node(11);
start.next.next.next.next = new node(12);
start.next.next.next.next.next = new node(11);
start.next.next.next.next.next.next = new node(10);
Console.WriteLine( "Linked list before removing "
+ "duplicates :" );
printList(start);
removeDuplicate(start);
Console.WriteLine( "\nLinked list after removing "
+ "duplicates :" );
printList(start);
}
}
|
Javascript
class node {
constructor(val) {
this .val = val;
this .next = null ;
}
}
function removeDuplicate( head) {
var hs = new Set();
var current = head;
var prev = null ;
while (current != null ) {
var curval = current.val;
if (hs.has(curval)) {
prev.next = current.next;
} else {
hs.add(curval);
prev = current;
}
current = current.next;
}
}
function printList( head) {
while (head != null ) {
console.log(head.val + " " );
head = head.next;
}
}
start = new node(10);
start.next = new node(12);
start.next.next = new node(11);
start.next.next.next = new node(11);
start.next.next.next.next = new node(12);
start.next.next.next.next.next = new node(11);
start.next.next.next.next.next.next = new node(10);
console.log(
"Linked list before removing duplicates :<br/>"
);
printList(start);
removeDuplicate(start);
console.log(
"<br/>Linked list after removing duplicates :<br/>"
);
printList(start);
|
Output
Linked list before removing duplicates :
10 12 11 11 12 11 10
Linked list after removing duplicates :
10 12 11
Time Complexity: O(N), on average (assuming that hash table access time is O(1) on average).
Auxiliary Space: O(N), As extra space is used to store the elements in the stack.
Last Updated :
12 Sep, 2023
Like Article
Save Article
Share your thoughts in the comments
Please Login to comment...