Remove duplicates from a sorted linked list (original) (raw)

Last Updated : 15 Apr, 2026

Given a linked list sorted in non-decreasing order.Return the list bydeleting 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 list

**Input : _Linked List = 5->10->10->20
**Output : 5->10->20 (After removing duplicate elements)'

Try It Yourself

Table of Content

• [Naive Approach] Using Hash Set – O(n) Time and O(n) Space
• [Expected Approach] By Changing Next Pointer – O(n) Time and O(1) Space
• [Alternate Approach] Using Recursion – O(n) Time and O(n) Space

**[Naive Approach] Using Hash Set – 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 Hash Set 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****.** C++ `

//Driver Code Starts #include using namespace std;

//Driver Code Ends

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;
}

// delete old list
curr = head;
while (curr != nullptr) {
    Node* temp = curr;
    curr = curr->next;
    delete temp;
}

// Return the head of the new list with
// duplicates removed
return new_head;

}

//Driver Code Starts

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);

printList(head);

head = removeDuplicates(head);

printList(head);

return 0;

} //Driver Code Ends

Java

//Driver Code Starts import java.io.*; import java.util.HashSet;

//Driver Code Ends

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();
}

//Driver Code Starts

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);

    printList(head);

    head = removeDuplicates(head);

    printList(head);
}

} //Driver Code Ends

Python

class Node: def init(self, x): self.data = x self.next = None

def removeDuplicates(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_list(head)

head = removeDuplicates(head)

print_list(head)

C#

//Driver Code Starts using System; using System.Collections.Generic;

//Driver Code Ends

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();

}

//Driver Code Starts

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);

    PrintList(head);
  
    head = removeDuplicates(head);

    PrintList(head);
}

} //Driver Code Ends

JavaScript

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);

printList(head);

head = removeDuplicates(head);

printList(head);

`

Output

11 11 11 13 13 20 11 13 20

**[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. C++ `

#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 *temp = curr->next;
        curr->next = curr->next->next;
        delete temp;
    }
    else
        curr = curr->next;
}
return head;

}

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);


// Original list
Node* temp = head;
while (temp != NULL) {
    cout << temp->data << " ";
    temp = temp->next;
}
cout << endl;

head = removeDuplicates(head);

// list withn no duplicates
temp = head;
while (temp != NULL) {
    cout << temp->data << " ";
    temp = temp->next;
}

return 0;

}

C

#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* temp = curr->next;
        curr->next = curr->next->next;
        free(temp);
    }
    else
        curr = curr->next;
}
return head;

}

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);

// original list
struct Node *temp = head;
while (temp != NULL) {
    printf("%d ", temp->data);
    temp = temp->next;
}
printf("\n");

head = removeDuplicates(head);

// list with no duplicates
temp = head;
while (temp != NULL) {
    printf("%d ", temp->data);
    temp = temp->next;
}
printf("\n");

return 0;

}

Java

//Driver Code Starts import java.io.*;

//Driver Code Ends

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;
}

// Driver code
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);

    
    Node temp = head;
    while (temp != null) {
        System.out.print(temp.data + " ");
        temp = temp.next;
    }
    System.out.println();

    head = removeDuplicates(head);

    // listwith no duplicate

//Driver Code Starts temp = head; while (temp != null) { System.out.print(temp.data + " "); temp = temp.next; } } } //Driver Code Ends

Python

class Node: def init(self, x): self.data = x self.next = None

def removeDuplicates(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

Driver code

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)

# original list
temp = head
while temp:
    print(temp.data, end=" ")
    temp = temp.next
print()

head = removeDuplicates(head)

# list with no duplicate
temp = head
while temp:
    print(temp.data, end=" ")
    temp = temp.next
print()

C#

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;
}

// Driver code
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);

    // original list
    Node temp = head;
    while (temp != null) {
        Console.Write(temp.data + " ");
        temp = temp.next;
    }
    Console.WriteLine();

    head = removeDuplicates(head);
    
    // list with no duplicates
    temp = head;
    while (temp != null) {
        Console.Write(temp.data + " ");
        temp = temp.next;
    }
    Console.WriteLine();
}

}

JavaScript

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;

}

// Driver code // 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);

// original list let temp = head; while (temp !== null) { process.stdout.write(temp.data + " "); temp = temp.next; } console.log();

head = removeDuplicates(head);

// list with no duplicates temp = head; while (temp !== null) { process.stdout.write(temp.data + " "); temp = temp.next; }

`

Output

11 11 11 13 13 20 11 13 20

**[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 nodeand 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. C++ `

//Driver Code Starts #include using namespace std;

//Driver Code Ends

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) {
        Node* temp = head->next;
        head->next = head->next->next;
        delete temp;
        removeDuplicates(head);
    }
    else{
        removeDuplicates(head->next);
    }
}

}

int main() {

// Create a sorted linked list

//Driver Code Starts // 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);

// printing old list
Node* temp = head;
while (temp != NULL) {
    cout << temp->data << " ";
    temp = temp->next;
}
cout << endl;

removeDuplicates(head);

// printing new list
temp = head;
while (temp != NULL) {
    cout << temp->data << " ";
    temp = temp->next;
}

return 0;

} //Driver Code Ends

C

#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) {
        struct Node* temp = head->next;
        head->next = head->next->next;
        free(temp); 
        // head->next = head->next->next;
        removeDuplicates(head);
    }
    else
        removeDuplicates(head->next);
}

}

// Driver code 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);

struct Node *temp = head;
while (temp != NULL) {
    printf("%d ", temp->data);
    temp = temp->next;
}
printf("\n");

removeDuplicates(head);

temp = head;
while (temp != NULL) {
    printf("%d ", temp->data);
    temp = temp->next;
}

return 0;

}

Java

//Driver Code Starts import java.io.*;

//Driver Code Ends

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);
        }
    }
}

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);

//Driver Code Starts head.next.next.next = new Node(13); head.next.next.next.next = new Node(13); head.next.next.next.next.next = new Node(20);

    // original list
    Node temp = head;
    while (temp != null) {
        System.out.print(temp.data + " ");
        temp = temp.next;
    }
    System.out.println();

    removeDuplicates(head);

    // list with no duplicates
    temp = head;
    while (temp != null) {
        System.out.print(temp.data + " ");
        temp = temp.next;
    }
}

} //Driver Code Ends

Python

class Node: def init(self, x): self.data = x self.next = None

def removeDuplicates(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
        removeDuplicates(head)
    else:

         # Continue with next node
        removeDuplicates(head.next)

Driver code

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)

# original list
temp = head
while temp:
    print(temp.data, end=" ")
    temp = temp.next
print()

removeDuplicates(head)

# list with no duplicates
temp = head
while temp:
    print(temp.data, end=" ")
    temp = temp.next
print()

C#

//Driver Code Starts using System;

//Driver Code Ends

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);
        }
    }
}

// Driver code
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);

//Driver Code Starts head.next.next.next.next = new Node(13); head.next.next.next.next.next = new Node(20);

    // original list
    Node temp = head;
    while (temp != null) {
        Console.Write(temp.data + " ");
        temp = temp.next;
    }
    Console.WriteLine();

    RemoveDuplicates(head);
    
    // list with no duplicates
    temp = head;
    while (temp != null) {
        Console.Write(temp.data + " ");
        temp = temp.next;
    }
}

} //Driver Code Ends

JavaScript

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);
    }
}

}

// Driver code // 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);

// original list let temp = head; while (temp !== null) { process.stdout.write(temp.data + " "); temp = temp.next; } console.log();

removeDuplicates(head);

// list with no duplicates temp = head; while (temp !== null) { process.stdout.write(temp.data + " "); temp = temp.next; } console.log();

`

Output

11 11 11 13 13 20 11 13 20