Remove duplicates from a sorted linked list
Last Updated :
26 Mar, 2025
Given a linked list sorted in non-decreasing order. Return the list by deleting the duplicate nodes from the list. The returned list should also be in non-decreasing order.
Example:
Input : Linked List = 11->11->11->21->43->43->60
Output : 11->21->43->60
Explanation:
Remove duplicates from a sorted linked listInput : Linked List = 5->10->10->20
Output : 5->10->20 (After removing duplicate elements)
[Naive Approach] Remove duplicates using HashSet – O(n) Time and O(n) Space:
The idea is to traverse the linked list and check if the value is present in HashSet or not. If the value is not present in the HashSet then push the value in HashSet append the nodes in the new list , otherwise skip the value as it is the duplicate value.
Follow the steps below to solve the problem:
- Initialize an empty HashSet and pointers new_head and tail as NULL.
- Iterate through the original list, adding each unique node's value to the HashSet and appending the node to the new list.
- Return the new_head of the new list with duplicates removed.
Below is the implementation of the above approach:
C++
// C++ Program to remove duplicates from a
// sorted linked list using Hashset
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int x) {
data = x;
next = nullptr;
}
};
Node *removeDuplicates(Node *head) {
// Unordered map to track unique node values
unordered_set<int>st;
// Initialize pointers for traversing the original list
// and building the new list without duplicates
Node *new_head = nullptr;
Node *tail = nullptr;
// Traverse the original list
Node *curr = head;
while (curr != nullptr) {
// Check if the current node's data is not in the map
if (st.find(curr->data) == st.end()) {
// Create a new node for the unique data
Node *new_node = new Node(curr->data);
// If new_head is null, this is the
// first unique node
if (new_head == nullptr) {
new_head = new_node;
tail = new_head;
}
else {
// Append the new node to the end
// of the new list
tail->next = new_node;
tail = new_node;
}
// Mark this data as encountered
st.insert(curr->data);
}
// Move to the next node in the original list
curr = curr->next;
}
// Return the head of the new list with
// duplicates removed
return new_head;
}
void printList(Node *node) {
while (node != NULL) {
cout << node->data << " ";
node = node->next;
}
cout << endl;
}
int main() {
// Create a sorted linked list
// 11->11->11->13->13->20
Node *head = new Node(11);
head->next = new Node(11);
head->next->next = new Node(11);
head->next->next->next = new Node(13);
head->next->next->next->next = new Node(13);
head->next->next->next->next->next = new Node(20);
cout << "Linked list before duplicate removal:" << endl;
printList(head);
head = removeDuplicates(head);
cout << "Linked list after duplicate removal:" << endl;
printList(head);
return 0;
}
// Java program to remove duplicates from a
// sorted linked list using Hashset
import java.io.*;
import java.util.HashSet;
class Node {
int data;
Node next;
Node(int x) {
data = x;
next = null;
}
}
class GfG {
// Function to remove duplicates
static Node removeDuplicates(Node head) {
// HashSet to track unique node values
HashSet<Integer> st = new HashSet<>();
// Initialize pointers for traversing
// the original list and building the new
// list without duplicates
Node temp = head;
Node newHead = null;
Node tail = null;
// Traverse the original list
while (temp != null) {
// Check if the current node's data is not in
// the set
if (!st.contains(temp.data)) {
// Create a new node for the unique data
Node newNode = new Node(temp.data);
// If newHead is null, this is the first
// unique node
if (newHead == null) {
newHead = newNode;
tail = newHead;
}
else {
// Append the new node to the
// end of the new list
tail.next = newNode;
tail = newNode;
}
// Mark this data as encountered
st.add(temp.data);
}
// Move to the next node in the original list
temp = temp.next;
}
// Return the head of the new list with
// duplicates removed
return newHead;
}
// Function to print nodes in a given linked list
public static void printList(Node node) {
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
System.out.println();
}
public static void main(String[] args) {
// Create a sorted linked list:
// 11->11->11->13->13->20
Node head = new Node(11);
head.next = new Node(11);
head.next.next = new Node(11);
head.next.next.next = new Node(13);
head.next.next.next.next = new Node(13);
head.next.next.next.next.next = new Node(20);
System.out.println(
"Linked list before duplicate removal:");
printList(head);
head = removeDuplicates(head);
System.out.println(
"Linked list after duplicate removal:");
printList(head);
}
}
# Python3 program to remove duplicate
# nodes from a sorted linked list using Hashset
class Node:
def __init__(self, x):
self.data = x
self.next = None
def remove_duplicates(head):
# Set to track unique node values
st = set()
# Initialize pointers for traversing
# the original list and building the
# new list without duplicates
temp = head
new_head = None
tail = None
# Traverse the original list
while temp:
# Check if the current node's data
# is not in the set
if temp.data not in st:
# Create a new node for the unique data
new_node = Node(temp.data)
# If new_head is None, this is the
#first unique node
if new_head is None:
new_head = new_node
tail = new_head
else:
# Append the new node to the end
#of the new list
tail.next = new_node
tail = new_node
# Mark this data as encountered
st.add(temp.data)
# Move to the next node in the original list
temp = temp.next
# Return the head of the new list with
#duplicates removed
return new_head
def print_list(node):
while node:
print(node.data, end=" ")
node = node.next
print()
if __name__ == "__main__":
# Create a sorted linked list:
# 11->11->11->13->13->20
head = Node(11)
head.next = Node(11)
head.next.next = Node(11)
head.next.next.next = Node(13)
head.next.next.next.next = Node(13)
head.next.next.next.next.next = Node(20)
print("Linked list before duplicate removal:")
print_list(head)
head = remove_duplicates(head)
print("Linked list after duplicate removal:")
print_list(head)
// C# program to remove duplicates
// from a sorted linked list using hashSet
using System;
using System.Collections.Generic;
public class Node {
public int data;
public Node next;
public Node(int x) {
data = x;
next = null;
}
}
class GfG {
// Function to remove duplicates
static Node RemoveDuplicates(Node head) {
// HashSet to track unique node values
HashSet<int> st = new HashSet<int>();
// Initialize pointers for traversing the original
// list and building the new list without duplicates
Node temp = head;
Node newHead = null;
Node tail = null;
// Traverse the original list
while (temp != null) {
// Check if the current node's data
// is not in the set
if (!st.Contains(temp.data)) {
// Create a new node for the unique data
Node newNode = new Node(temp.data);
// If newHead is null, this is the
// first unique node
if (newHead == null) {
newHead = newNode;
tail = newHead;
}
else {
// Append the new node to the end
// of the new list
tail.next = newNode;
tail = newNode;
}
// Mark this data as encountered
st.Add(temp.data);
}
// Move to the next node in the
//original list
temp = temp.next;
}
// Return the head of the new list
return newHead;
}
static void PrintList(Node node) {
while (node != null) {
Console.Write(node.data + " ");
node = node.next;
}
Console.WriteLine();
}
static void Main() {
// Create a sorted linked list:
//11->11->11->13->13->20
Node head = new Node(11);
head.next = new Node(11);
head.next.next = new Node(11);
head.next.next.next = new Node(13);
head.next.next.next.next = new Node(13);
head.next.next.next.next.next = new Node(20);
Console.WriteLine(
"Linked list before duplicate removal:");
PrintList(head);
head = RemoveDuplicates(head);
Console.WriteLine(
"Linked list after duplicate removal:");
PrintList(head);
}
}
// JavaScript program to remove duplicates
// from a sorted linked list using hashSet
class Node {
constructor(x) {
this.data = x;
this.next = null;
}
}
function removeDuplicates(head) {
// Set to track unique node values
const st = new Set();
// Initialize pointers for traversing the original list
// and building the new list without duplicates
let temp = head;
let newHead = null;
let tail = null;
// Traverse the original list
while (temp !== null) {
// Check if the current node's
// data is not in the set
if (!st.has(temp.data)) {
// Create a new node for the unique data
const newNode = new Node(temp.data);
// If newHead is null, this is the first
// unique node
if (newHead === null) {
newHead = newNode;
tail = newHead;
}
else {
// Append the new node to the end of
// the new list
tail.next = newNode;
tail = newNode;
}
// Mark this data as encountered
st.add(temp.data);
}
// Move to the next node in the original list
temp = temp.next;
}
// Return the head of the new list with
// duplicates removed
return newHead;
}
function printList(node) {
let current = node;
while (current) {
process.stdout.write(current.data + " ");
current = current.next;
}
console.log();
}
// Create a sorted linked list:
// 11->11->11->13->13->20
let head = new Node(11);
head.next = new Node(11);
head.next.next = new Node(11);
head.next.next.next = new Node(13);
head.next.next.next.next = new Node(13);
head.next.next.next.next.next = new Node(20);
console.log("Linked list before duplicate removal:");
printList(head);
head = removeDuplicates(head);
console.log("Linked list after duplicate removal:");
printList(head);
OutputLinked list before duplicate removal:
11 11 11 13 13 20
Linked list after duplicate removal:
11 13 20
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(n)
[Expected Approach] By Changing Next Pointer – O(n) Time and O(1) Space:
The idea is to traverse the linked list and for each node, if the next node has the same data, skip and delete the duplicate node.
Follow the steps below to solve the problem:
- Traverse the linked list starting from the head node.
- Iterate through the list, comparing each node with the next node.
- If the data in the next node is same as the curr node adjust pointers to skip the next node.
Below is the implementation of the above approach:
C++
// C++ Program to remove duplicates from a
// sorted linked list
#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int x) {
data = x;
next = nullptr;
}
};
Node *removeDuplicates(Node *head) {
Node *curr = head;
// Traverse the list
while (curr != NULL && curr->next != NULL) {
// Check if next value is same as current
if (curr->data == curr->next->data) {
Node *next_next = curr->next->next;
curr->next = next_next;
}
else
curr = curr->next;
}
return head;
}
void printList(Node *node) {
while (node != NULL) {
cout << node->data << " ";
node = node->next;
}
cout << endl;
}
int main() {
// Create a sorted linked list
// 11->11->11->13->13->20
Node *head = new Node(11);
head->next = new Node(11);
head->next->next = new Node(11);
head->next->next->next = new Node(13);
head->next->next->next->next = new Node(13);
head->next->next->next->next->next = new Node(20);
cout << "Linked list before duplicate removal:" << endl;
printList(head);
head = removeDuplicates(head);
cout << "Linked list after duplicate removal:" << endl;
printList(head);
return 0;
}
// C Program to remove duplicates from a
// sorted linked list
#include <stdio.h>
struct Node {
int data;
struct Node *next;
};
// Function to remove duplicates
struct Node *removeDuplicates(struct Node *head) {
struct Node *curr = head;
// Traverse the list
while (curr != NULL && curr->next != NULL) {
// Check if next value is the same as curr
if (curr->data == curr->next->data) {
struct Node *next_next = curr->next->next;
curr->next = next_next;
}
else
curr = curr->next;
}
return head;
}
void printList(struct Node *node) {
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
printf("\n");
}
struct Node *createNode(int new_data){
struct Node *new_node =
(struct Node *)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = NULL;
return new_node;
}
int main() {
// Create a sorted linked list:
// 11->11->11->13->13->20
struct Node *head = createNode(11);
head->next = createNode(11);
head->next->next = createNode(11);
head->next->next->next = createNode(13);
head->next->next->next->next = createNode(13);
head->next->next->next->next->next = createNode(20);
printf("Linked list before duplicate removal:\n");
printList(head);
head = removeDuplicates(head);
printf("Linked list after duplicate removal:\n");
printList(head);
return 0;
}
// Java program to remove duplicates
// from a sorted linked list
import java.io.*;
class Node {
int data;
Node next;
Node(int x) {
data = x;
next = null;
}
}
class GfG {
// Function to remove duplicates
static Node removeDuplicates(Node head) {
Node curr = head;
// Traverse the list
while (curr != null && curr.next != null) {
// Check if next value is the same as curr
if (curr.data == curr.next.data) {
Node nextNext = curr.next.next;
curr.next = nextNext;
}
else {
curr = curr.next;
}
}
return head;
}
static void printList(Node node) {
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
System.out.println();
}
public static void main(String[] args) {
// Create a sorted linked list:
// 11->11->11->13->13->20
Node head = new Node(11);
head.next = new Node(11);
head.next.next = new Node(11);
head.next.next.next = new Node(13);
head.next.next.next.next = new Node(13);
head.next.next.next.next.next = new Node(20);
System.out.println(
"Linked list before duplicate removal:");
printList(head);
head = removeDuplicates(head);
System.out.println(
"Linked list after duplicate removal:");
printList(head);
}
}
# Python3 program to remove duplicate
# nodes from a sorted linked list
class Node:
def __init__(self, x):
self.data = x
self.next = None
def remove_duplicates(head):
curr = head
# Traverse the list
while curr and curr.next:
# Check if next value is the same as curr
if curr.data == curr.next.data:
next_next = curr.next.next
curr.next = next_next
else:
curr = curr.next
return head
def print_list(node):
while node:
print(node.data, end=" ")
node = node.next
print()
if __name__ == "__main__":
# Create a sorted linked list:
# 11->11->11->13->13->20
head = Node(11)
head.next = Node(11)
head.next.next = Node(11)
head.next.next.next = Node(13)
head.next.next.next.next = Node(13)
head.next.next.next.next.next = Node(20)
print("Linked list before duplicate removal:")
print_list(head)
head = remove_duplicates(head)
print("Linked list after duplicate removal:")
print_list(head)
// C# program to remove duplicates
// from a sorted linked list
using System;
public class Node {
public int data;
public Node next;
public Node(int x) {
data = x;
next = null;
}
}
class GfG {
// Function to remove duplicates
static Node RemoveDuplicates(Node head) {
Node curr = head;
// Traverse the list
while (curr != null && curr.next != null) {
// Check if next value is the same as curr
if (curr.data == curr.next.data) {
Node nextNext = curr.next.next;
curr.next = nextNext;
}
else {
curr = curr.next;
}
}
return head;
}
static void PrintList(Node node) {
while (node != null) {
Console.Write(node.data + " ");
node = node.next;
}
Console.WriteLine();
}
static void Main() {
// Create a sorted linked list:
// 11->11->11->13->13->20
Node head = new Node(11);
head.next = new Node(11);
head.next.next = new Node(11);
head.next.next.next = new Node(13);
head.next.next.next.next = new Node(13);
head.next.next.next.next.next = new Node(20);
Console.WriteLine(
"Linked list before duplicate removal:");
PrintList(head);
head = RemoveDuplicates(head);
Console.WriteLine(
"Linked list after duplicate removal:");
PrintList(head);
}
}
// JavaScript program to remove duplicates
// from a sorted linked list
class Node {
constructor(x) {
this.data = x;
this.next = null;
}
}
function removeDuplicates(head) {
let curr = head;
// Traverse the list
while (curr && curr.next) {
// Check if next value is the same as curr
if (curr.data === curr.next.data) {
let nextNext = curr.next.next;
curr.next = nextNext;
}
else {
curr = curr.next;
}
}
return head;
}
function printList(node) {
let current = node;
while (current) {
process.stdout.write(current.data + " ");
current = current.next;
}
console.log();
}
// Create a sorted linked list:
// 11->11->11->13->13->20
let head = new Node(11);
head.next = new Node(11);
head.next.next = new Node(11);
head.next.next.next = new Node(13);
head.next.next.next.next = new Node(13);
head.next.next.next.next.next = new Node(20);
console.log("Linked list before duplicate removal:");
printList(head);
head = removeDuplicates(head);
console.log("Linked list after duplicate removal:");
printList(head);
OutputLinked list before duplicate removal:
11 11 11 13 13 20
Linked list after duplicate removal:
11 13 20
Time Complexity: O(n) where n is the number of nodes in the given linked list.
Auxiliary Space: O(1)
[Alternate Approach] Using Recursion – O(n) Time and O(n) Space:
The idea is similar to the iterative approach. Here we are using the recursion to check each node and its next for duplicates. Please note that the iterative approach would be better in terns of time and space. The recursive approach can be good fun exercise or a question in an interview / exam.
Follow the steps below to solve the problem:
- If the curr node or its next node is NULL, return the curr node.
- If the current node’s data equals the next node’s data, adjust pointers to skip the duplicate.
- If no duplicate, recursively process the next node.
Below is the implementation of the above approach:
C++
// C++ Program to remove duplicates from a
// sorted linked list
#include <iostream>
using namespace std;
class Node {
public:
int data;
Node *next;
Node(int new_data) {
data = new_data;
next = nullptr;
}
};
// Function to remove duplicates
void removeDuplicates(Node *head) {
// Base case: if the list is empty, return
if (head == NULL)
return;
// Check if the next node exists
if (head->next != NULL) {
// If current node has duplicate
// data with the next node
if (head->data == head->next->data) {
head->next = head->next->next;
removeDuplicates(head);
}
else{
removeDuplicates(head->next);
}
}
}
void printList(Node *node) {
while (node != NULL) {
cout << node->data << " ";
node = node->next;
}
cout << endl;
}
int main() {
// Create a sorted linked list
// 11->11->11->13->13->20
Node *head = new Node(11);
head->next = new Node(11);
head->next->next = new Node(11);
head->next->next->next = new Node(13);
head->next->next->next->next = new Node(13);
head->next->next->next->next->next = new Node(20);
cout << "Linked list before duplicate removal:" << endl;
printList(head);
removeDuplicates(head);
cout << "Linked list after duplicate removal:" << endl;
printList(head);
return 0;
}
// C Program to remove duplicates from a
// sorted linked list
#include <stdio.h>
struct Node {
int data;
struct Node *next;
};
struct Node *createNode(int new_data) {
struct Node *new_node =
(struct Node *)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = NULL;
return new_node;
}
// Function to remove duplicates
void removeDuplicates(struct Node *head) {
// Base case: if the list is empty, return
if (head == NULL)
return;
// Check if the next node exists
if (head->next != NULL) {
// If current node has duplicate data
// with the next node
if (head->data == head->next->data) {
head->next = head->next->next;
removeDuplicates(head);
}
else
removeDuplicates(head->next);
}
}
void printList(struct Node *node) {
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
printf("\n");
}
int main() {
// Create a sorted linked list:
// 11->11->11->13->13->20
struct Node *head = createNode(11);
head->next = createNode(11);
head->next->next = createNode(11);
head->next->next->next = createNode(13);
head->next->next->next->next = createNode(13);
head->next->next->next->next->next = createNode(20);
printf("Linked list before duplicate removal:\n");
printList(head);
removeDuplicates(head);
printf("Linked list after duplicate removal:\n");
printList(head);
return 0;
}
// Java program to remove duplicates
// from a sorted linked list
import java.io.*;
class Node {
int data;
Node next;
Node(int x) {
data = x;
next = null;
}
}
class GfG {
// Function to remove duplicates
static void removeDuplicates(Node head) {
// Base case: if the list is empty, return
if (head == null)
return;
// Check if the next node exists
if (head.next != null) {
// If current node has duplicate data with the
// next node
if (head.data == head.next.data) {
head.next = head.next.next;
removeDuplicates(head);
}
else {
// Continue with next node
removeDuplicates(head.next);
}
}
}
static void printList(Node node) {
while (node != null) {
System.out.print(node.data + " ");
node = node.next;
}
System.out.println();
}
public static void main(String[] args) {
// Create a sorted linked list:
// 11->11->11->13->13->20
Node head = new Node(11);
head.next = new Node(11);
head.next.next = new Node(11);
head.next.next.next = new Node(13);
head.next.next.next.next = new Node(13);
head.next.next.next.next.next = new Node(20);
System.out.println(
"Linked list before duplicate removal:");
printList(head);
removeDuplicates(head);
System.out.println(
"Linked list after duplicate removal:");
printList(head);
}
}
# Python3 program to remove duplicate
# nodes from a sorted linked list
class Node:
def __init__(self, x):
self.data = x
self.next = None
def remove_duplicates(head):
# Base case: if the list is empty, return
if head is None:
return
# Check if the next node exists
if head.next is not None:
# If current node has duplicate data with the next node
if head.data == head.next.data:
head.next = head.next.next
remove_duplicates(head)
else:
# Continue with next node
remove_duplicates(head.next)
def print_list(node):
while node:
print(node.data, end=" ")
node = node.next
print()
if __name__ == "__main__":
# Create a sorted linked list:
# 11->11->11->13->13->20
head = Node(11)
head.next = Node(11)
head.next.next = Node(11)
head.next.next.next = Node(13)
head.next.next.next.next = Node(13)
head.next.next.next.next.next = Node(20)
print("Linked list before duplicate removal:")
print_list(head)
remove_duplicates(head)
print("Linked list after duplicate removal:")
print_list(head)
// C# program to remove duplicates from a sorted linked list
using System;
public class Node {
public int data;
public Node next;
public Node(int x) {
data = x;
next = null;
}
}
class GfG {
// Function to remove duplicates
static void RemoveDuplicates(Node head) {
// Base case: if the list is empty, return
if (head == null)
return;
// Check if the next node exists
if (head.next != null) {
// If current node has duplicate data with
// the next node
if (head.data == head.next.data) {
head.next = head.next.next;
RemoveDuplicates(head);
}
else {
// Continue with next node
RemoveDuplicates(head.next);
}
}
}
static void PrintList(Node node) {
while (node != null) {
Console.Write(node.data + " ");
node = node.next;
}
Console.WriteLine();
}
static void Main() {
// Create a sorted linked list:
// 11->11->11->13->13->20
Node head = new Node(11);
head.next = new Node(11);
head.next.next = new Node(11);
head.next.next.next = new Node(13);
head.next.next.next.next = new Node(13);
head.next.next.next.next.next = new Node(20);
Console.WriteLine(
"Linked list before duplicate removal:");
PrintList(head);
RemoveDuplicates(head);
Console.WriteLine(
"Linked list after duplicate removal:");
PrintList(head);
}
}
// JavaScript program to remove duplicates from a sorted
// linked list
class Node {
constructor(x) {
this.data = x;
this.next = null;
}
}
function removeDuplicates(head) {
// Base case: if the list is empty, return
if (head === null)
return;
// Check if the next node exists
if (head.next !== null) {
// If current node has duplicate data
// with the next node
if (head.data === head.next.data) {
head.next = head.next.next;
removeDuplicates(head);
}
else {
// Continue with next node
removeDuplicates(head.next);
}
}
}
function printList(node) {
let current = node;
while (current) {
process.stdout.write(current.data + " ");
current = current.next;
}
console.log();
}
// Create a sorted linked list:
// 11->11->11->13->13->20
let head = new Node(11);
head.next = new Node(11);
head.next.next = new Node(11);
head.next.next.next = new Node(13);
head.next.next.next.next = new Node(13);
head.next.next.next.next.next = new Node(20);
console.log("Linked list before duplicate removal:");
printList(head);
removeDuplicates(head);
console.log("Linked list after duplicate removal:");
printList(head);
OutputLinked list before duplicate removal:
11 11 11 13 13 20
Linked list after duplicate removal:
11 13 20
Time Complexity: O(n) , where n is the number of nodes in the given linked list.
Auxiliary Space: O(n)
Remove duplicates from a sorted linked list
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