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 list
Input : 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)