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Merge two sorted linked list without duplicates

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  • Difficulty Level : Easy
  • Last Updated : 30 Aug, 2022
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Merge two sorted linked list of size n1 and n2. The duplicates in two linked list should be present only once in the final sorted linked list.

Examples:  

Input : list1: 1->1->4->5->7
        list2: 2->4->7->9
Output : 1 2 4 5 7 9

Source: Microsoft on Campus Placement and Interview Questions

Approach: Following are the steps: 

  1. Merge the two sorted linked list in sorted manner. Refer recursive approach of this post. Let the final obtained list be head.
  2. Remove duplicates from sorted linked list head.

Implementation:

C++




// C++ implementation to merge two sorted linked list
// without duplicates
#include <bits/stdc++.h>
 
using namespace std;
 
// structure of a node
struct Node {
    int data;
    Node* next;
};
 
// function to get a new node
Node* getNode(int data)
{
    // allocate space
    Node* temp = (Node*)malloc(sizeof(Node));
 
    // put in data
    temp->data = data;
    temp->next = NULL;
    return temp;
}
 
// function to merge two sorted linked list
// in a sorted manner
Node* sortedMerge(struct Node* a, struct Node* b)
{
    Node* result = NULL;
 
    /* Base cases */
    if (a == NULL)
        return (b);
    else if (b == NULL)
        return (a);
 
    /* Pick either a or b, and recur */
    if (a->data <= b->data) {
        result = a;
        result->next = sortedMerge(a->next, b);
    }
    else {
        result = b;
        result->next = sortedMerge(a, b->next);
    }
    return (result);
}
 
/* The function removes duplicates from a sorted list */
void removeDuplicates(Node* head)
{
    /* Pointer to traverse the linked list */
    Node* current = head;
 
    /* Pointer to store the next pointer of a node to be deleted*/
    Node* next_next;
 
    /* do nothing if the list is empty */
    if (current == NULL)
        return;
 
    /* Traverse the list till last node */
    while (current->next != NULL) {
 
        /* Compare current node with next node */
        if (current->data == current->next->data) {
 
            /* The sequence of steps is important*/
            next_next = current->next->next;
            free(current->next);
            current->next = next_next;
        }
        else /* This is tricky: only advance if no deletion */
        {
            current = current->next;
        }
    }
}
 
// function to merge two sorted linked list
// without duplicates
Node* sortedMergeWithoutDuplicates(Node* head1, Node* head2)
{
    // merge two linked list in sorted manner
    Node* head = sortedMerge(head1, head2);
 
    // remove duplicates from the list 'head'
    removeDuplicates(head);
 
    return head;
}
 
// function to print the linked list
void printList(Node* head)
{
    while (head != NULL) {
        cout << head->data << " ";
        head = head->next;
    }
}
 
// Driver program to test above
int main()
{
    // head1: 1->1->4->5->7
    Node* head1 = getNode(1);
    head1->next = getNode(1);
    head1->next->next = getNode(4);
    head1->next->next->next = getNode(5);
    head1->next->next->next->next = getNode(7);
 
    // head2: 2->4->7->9
    Node* head2 = getNode(2);
    head2->next = getNode(4);
    head2->next->next = getNode(7);
    head2->next->next->next = getNode(9);
 
    Node* head3;
 
    head3 = sortedMergeWithoutDuplicates(head1, head2);
 
    printList(head3);
 
    return 0;
}

Java




// Java implementation to merge two sorted linked list
// without duplicates
import java.io.*;
 
class GFG {
 
    // structure of a node
    class Node {
        int data;
        Node next;
    }
 
    // function to get a new node
    public Node getNode(int data)
    {
        Node temp = new Node();
        temp.data = data;
        temp.next = null;
        return temp;
    }
 
    // function to merge two sorted linked list in a sorted
    // manner
    static Node sortedMerge(Node a, Node b)
    {
        Node result = null;
        /* Base cases */
        if (a == null) {
            return b;
        }
        else if (b == null) {
            return a;
        }
        /* Pick either a or b, and recur */
        if (a.data <= b.data) {
            result = a;
            result.next = sortedMerge(a.next, b);
        }
        else {
            result = b;
            result.next = sortedMerge(a, b.next);
        }
        return result;
    }
 
    /* The function removes duplicates from a sorted list */
    static void removeDuplicates(Node head)
    {
        /* Pointer to traverse the linked list */
        Node current = head;
       
        /* Pointer to store the next pointer of a node to be
         * deleted*/
        Node next_next;
 
        /* do nothing if the list is empty */
        if (current == null) {
            return;
        }
 
        /* Traverse the list till last node */
        while (current.next != null)
        {
           
            /* Compare current node with next node */
            if (current.data == current.next.data)
            {
               
                /* The sequence of steps is important*/
                next_next = current.next.next;
                current.next = next_next;
            }
            else { /* This is tricky: only advance if no
                      deletion */
                current = current.next;
            }
        }
    }
 
    // function to merge two sorted linked list without
    // duplicates
    public Node sortedMergeWithoutDuplicates(Node head1,
                                             Node head2)
    {
 
        // merge two linked list in sorted manner
        Node head = sortedMerge(head1, head2);
 
        // remove duplicates from the list 'head'
        removeDuplicates(head);
 
        return head;
    }
 
    // function to print the linked list
    public void printList(Node head)
    {
        while (head != null) {
            System.out.print(head.data + " ");
            head = head.next;
        }
    }
 
    public static void main(String[] args)
    {
 
        GFG l = new GFG();
 
        // head1 : 1->1->4->5->7
        Node head1 = l.getNode(1);
        head1.next = l.getNode(1);
        head1.next.next = l.getNode(4);
        head1.next.next.next = l.getNode(5);
        head1.next.next.next.next = l.getNode(7);
 
        // head2 : 2->4->7->9
        Node head2 = l.getNode(2);
        head2.next = l.getNode(4);
        head2.next.next = l.getNode(7);
        head2.next.next.next = l.getNode(9);
 
        Node head3;
 
        head3
            = l.sortedMergeWithoutDuplicates(head1, head2);
 
        l.printList(head3);
    }
}
 
// This code is contributed by lokeshmvs21.

Python3




# Python3 implementation to merge two
# sorted linked list without duplicates
  
# Structure of a node
class Node:
     
    def __init__(self, data):
         
        self.data = data
        self.next = None
  
# Function to get a new node
def getNode(data):
     
    # Allocate space
    temp = Node(data)
    return temp
 
# Function to merge two sorted linked
# list in a sorted manner
def sortedMerge(a, b):
     
    result = None
  
    # Base cases
    if (a == None):
        return(b)
    elif (b == None):
        return(a)
  
    # Pick either a or b, and recur
    if (a.data <= b.data):
        result = a
        result.next = sortedMerge(a.next, b)
    else:
        result = b
        result.next = sortedMerge(a, b.next)
 
    return(result)
     
# The function removes duplicates
# from a sorted list
def removeDuplicates(head):
 
    # Pointer to traverse the linked list
    current = head
  
    # Pointer to store the next pointer
    # of a node to be deleted
    next_next = None
  
    # Do nothing if the list is empty
    if (current == None):
        return
  
    # Traverse the list till last node
    while (current.next != None):
  
        # Compare current node with next node
        if (current.data == current.next.data):
  
            # The sequence of steps is important
            next_next = current.next.next
            del (current.next)
            current.next = next_next
        else:
             
            # This is tricky: only advance
            # if no deletion
            current = current.next
     
# Function to merge two sorted linked list
# without duplicates
def sortedMergeWithoutDuplicates(head1, head2):
 
    # Merge two linked list in sorted manner
    head = sortedMerge(head1, head2)
  
    # Remove duplicates from the list 'head'
    removeDuplicates(head)
  
    return head
 
# Function to print the linked list
def printList(head):
 
    while (head != None):
        print(head.data, end = ' ')   
        head = head.next
     
# Driver code
if __name__=='__main__':
     
    # head1: 1.1.4.5.7
    head1 = getNode(1)
    head1.next = getNode(1)
    head1.next.next = getNode(4)
    head1.next.next.next = getNode(5)
    head1.next.next.next.next = getNode(7)
  
    # head2: 2.4.7.9
    head2 = getNode(2)
    head2.next = getNode(4)
    head2.next.next = getNode(7)
    head2.next.next.next = getNode(9)
  
    head3 = sortedMergeWithoutDuplicates(
        head1, head2)
  
    printList(head3)
  
# This code is contributed by rutvik_56

C#




// C# implementation to merge two sorted linked list
// without duplicates
using System;
public class GFG{
 
  // structure of a node
  class Node {
    public int data;
    public Node next;
  }
 
  // function to get a new node
  Node getNode(int data)
  {
    Node temp = new Node();
    temp.data = data;
    temp.next = null;
    return temp;
  }
 
  // function to merge two sorted linked list in a sorted
  // manner
  static Node sortedMerge(Node a, Node b)
  {
    Node result = null;
    /* Base cases */
    if (a == null) {
      return b;
    }
    else if (b == null) {
      return a;
    }
    /* Pick either a or b, and recur */
    if (a.data <= b.data) {
      result = a;
      result.next = sortedMerge(a.next, b);
    }
    else {
      result = b;
      result.next = sortedMerge(a, b.next);
    }
    return result;
  }
 
  /* The function removes duplicates from a sorted list */
  static void removeDuplicates(Node head)
  {
    /* Pointer to traverse the linked list */
    Node current = head;
 
    /* Pointer to store the next pointer of a node to be
         * deleted*/
    Node next_next;
 
    /* do nothing if the list is empty */
    if (current == null) {
      return;
    }
 
    /* Traverse the list till last node */
    while (current.next != null)
    {
 
      /* Compare current node with next node */
      if (current.data == current.next.data)
      {
 
        /* The sequence of steps is important*/
        next_next = current.next.next;
        current.next = next_next;
      }
      else { /* This is tricky: only advance if no
                      deletion */
        current = current.next;
      }
    }
  }
 
  // function to merge two sorted linked list without
  // duplicates
  Node sortedMergeWithoutDuplicates(Node head1,
                                    Node head2)
  {
 
    // merge two linked list in sorted manner
    Node head = sortedMerge(head1, head2);
 
    // remove duplicates from the list 'head'
    removeDuplicates(head);
 
    return head;
  }
 
  // function to print the linked list
  void printList(Node head)
  {
    while (head != null) {
      Console.Write(head.data + " ");
      head = head.next;
    }
  }
 
  static public void Main (){
 
    GFG l = new GFG();
 
    // head1 : 1->1->4->5->7
    Node head1 = l.getNode(1);
    head1.next = l.getNode(1);
    head1.next.next = l.getNode(4);
    head1.next.next.next = l.getNode(5);
    head1.next.next.next.next = l.getNode(7);
 
    // head2 : 2->4->7->9
    Node head2 = l.getNode(2);
    head2.next = l.getNode(4);
    head2.next.next = l.getNode(7);
    head2.next.next.next = l.getNode(9);
 
    Node head3;
 
    head3
      = l.sortedMergeWithoutDuplicates(head1, head2);
 
    l.printList(head3);
  }
}
 
// This code is contributed by lokeshmvs21.

Output

1 2 4 5 7 9 

Complexity Analysis:

  • Time complexity: O(n1 + n2). 
  • Auxiliary Space: O(1).

Exercise: Get the final sorted linked list without duplicates in a single traversal of the two lists.


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