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Sorted merge of two sorted doubly circular linked lists

  • Difficulty Level : Medium
  • Last Updated : 12 May, 2021

Given two sorted Doubly circular Linked List containing n1 and n2 nodes respectively. The problem is to merge the two lists such that resultant list is also in sorted order.
Example: 
List 1: 
 

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List 2: 
 



Final list: 
 

 

Approach: Following are the steps:
 

  1. If head1 == NULL, return head2.
  2. If head2 == NULL, return head1.
  3. Let last1 and last2 be the last nodes of the two lists respectively. They can be obtained with the help of the previous links of the first nodes.
  4. Get pointer to the node which will be the last node of the final list. If last1.data < last2.data, then last_node = last2, Else last_node = last1.
  5. Update last1.next = last2.next = NULL.
  6. Now merge the two lists as two sorted doubly linked list are being merged. Refer merge procedure of this post. Let the first node of the final list be finalHead.
  7. Update finalHead.prev = last_node and last_node.next = finalHead.
  8. Return finalHead.

 

C++




// C++ implementation for Sorted merge of two
// sorted doubly circular linked list
#include <bits/stdc++.h>
using namespace std;
 
struct Node {
    int data;
    Node *next, *prev;
};
 
// A utility function to insert a new node at the
// beginning of doubly circular linked list
void insert(Node** head_ref, int data)
{
    // allocate space
    Node* new_node = new Node;
 
    // put in the data
    new_node->data = data;
 
    // if list is empty
    if (*head_ref == NULL) {
        new_node->next = new_node;
        new_node->prev = new_node;
    }
 
    else {
 
        // pointer points to last Node
        Node* last = (*head_ref)->prev;
 
        // setting up previous and next of new node
        new_node->next = *head_ref;
        new_node->prev = last;
 
        // update next and previous pointers of head_ref
        // and last.
        last->next = (*head_ref)->prev = new_node;
    }
 
    // update head_ref pointer
    *head_ref = new_node;
}
 
// function for Sorted merge of two
// sorted doubly linked list
Node* merge(Node* first, Node* second)
{
    // If first list is empty
    if (!first)
        return second;
 
    // If second list is empty
    if (!second)
        return first;
 
    // Pick the smaller value and adjust
    // the links
    if (first->data < second->data) {
        first->next = merge(first->next, second);
        first->next->prev = first;
        first->prev = NULL;
        return first;
    }
    else {
        second->next = merge(first, second->next);
        second->next->prev = second;
        second->prev = NULL;
        return second;
    }
}
 
// function for Sorted merge of two sorted
// doubly circular linked list
Node* mergeUtil(Node* head1, Node* head2)
{
    // if 1st list is empty
    if (!head1)
        return head2;
 
    // if 2nd list is empty
    if (!head2)
        return head1;
 
    // get pointer to the node which will be the
    // last node of the final list
    Node* last_node;
    if (head1->prev->data < head2->prev->data)
        last_node = head2->prev;
    else
        last_node = head1->prev;
 
    // store NULL to the 'next' link of the last nodes
    // of the two lists
    head1->prev->next = head2->prev->next = NULL;
 
    // sorted merge of head1 and head2
    Node* finalHead = merge(head1, head2);
 
    // 'prev' of 1st node pointing the last node
    // 'next' of last node pointing to 1st node
    finalHead->prev = last_node;
    last_node->next = finalHead;
 
    return finalHead;
}
 
// function to print the list
void printList(Node* head)
{
    Node* temp = head;
 
    while (temp->next != head) {
        cout << temp->data << " ";
        temp = temp->next;
    }
    cout << temp->data << " ";
}
 
// Driver program to test above
int main()
{
    Node *head1 = NULL, *head2 = NULL;
 
    // list 1:
    insert(&head1, 8);
    insert(&head1, 5);
    insert(&head1, 3);
    insert(&head1, 1);
 
    // list 2:
    insert(&head2, 11);
    insert(&head2, 9);
    insert(&head2, 7);
    insert(&head2, 2);
 
    Node* newHead = mergeUtil(head1, head2);
 
    cout << "Final Sorted List: ";
    printList(newHead);
 
    return 0;
}

Java




// Java implementation for Sorted merge of two
// sorted doubly circular linked list
class GFG
{
     
static class Node
{
    int data;
    Node next, prev;
};
 
// A utility function to insert a new node at the
// beginning of doubly circular linked list
static Node insert(Node head_ref, int data)
{
    // allocate space
    Node new_node = new Node();
 
    // put in the data
    new_node.data = data;
 
    // if list is empty
    if (head_ref == null)
    {
        new_node.next = new_node;
        new_node.prev = new_node;
    }
 
    else
    {
 
        // pointer points to last Node
        Node last = (head_ref).prev;
 
        // setting up previous and next of new node
        new_node.next = head_ref;
        new_node.prev = last;
 
        // update next and previous pointers of head_ref
        // and last.
        last.next = (head_ref).prev = new_node;
    }
 
    // update head_ref pointer
    head_ref = new_node;
    return head_ref;
}
 
// function for Sorted merge of two
// sorted doubly linked list
static Node merge(Node first, Node second)
{
    // If first list is empty
    if (first == null)
        return second;
 
    // If second list is empty
    if (second == null)
        return first;
 
    // Pick the smaller value and adjust
    // the links
    if (first.data < second.data)
    {
        first.next = merge(first.next, second);
        first.next.prev = first;
        first.prev = null;
        return first;
    }
    else
    {
        second.next = merge(first, second.next);
        second.next.prev = second;
        second.prev = null;
        return second;
    }
}
 
// function for Sorted merge of two sorted
// doubly circular linked list
static Node mergeUtil(Node head1, Node head2)
{
    // if 1st list is empty
    if (head1 == null)
        return head2;
 
    // if 2nd list is empty
    if (head2 == null)
        return head1;
 
    // get pointer to the node which will be the
    // last node of the final list
    Node last_node;
    if (head1.prev.data < head2.prev.data)
        last_node = head2.prev;
    else
        last_node = head1.prev;
 
    // store null to the 'next' link of the last nodes
    // of the two lists
    head1.prev.next = head2.prev.next = null;
 
    // sorted merge of head1 and head2
    Node finalHead = merge(head1, head2);
 
    // 'prev' of 1st node pointing the last node
    // 'next' of last node pointing to 1st node
    finalHead.prev = last_node;
    last_node.next = finalHead;
 
    return finalHead;
}
 
// function to print the list
static void printList(Node head)
{
    Node temp = head;
 
    while (temp.next != head)
    {
        System.out.print ( temp.data+ " ");
        temp = temp.next;
    }
    System.out.print ( temp.data + " ");
}
 
// Driver code
public static void main(String args[])
{
    Node head1 = null, head2 = null;
 
    // list 1:
    head1 = insert(head1, 8);
    head1 = insert(head1, 5);
    head1 = insert(head1, 3);
    head1 = insert(head1, 1);
 
    // list 2:
    head2 = insert(head2, 11);
    head2 = insert(head2, 9);
    head2 = insert(head2, 7);
    head2 = insert(head2, 2);
 
    Node newHead = mergeUtil(head1, head2);
 
    System.out.print( "Final Sorted List: ");
    printList(newHead);
}
}
 
// This code is contributed by Arnab Kundu

Python3




    # Python3 implementation for Sorted merge
# of two sorted doubly circular linked list
import math
 
class Node:
    def __init__(self, data):
        self.data = data
        self.next = None
        self.prev = None
 
# A utility function to insert
# a new node at the beginning
# of doubly circular linked list
def insert(head_ref, data):
     
    # allocate space
    new_node = Node(data)
 
    # put in the data
    new_node.data = data
 
    # if list is empty
    if (head_ref == None):
        new_node.next = new_node
        new_node.prev = new_node
     
    else :
 
        # pointer points to last Node
        last = head_ref.prev
 
        # setting up previous and
        # next of new node
        new_node.next = head_ref
        new_node.prev = last
 
        # update next and previous pointers
        # of head_ref and last.
        last.next = new_node
        head_ref.prev = new_node
     
    # update head_ref pointer
    head_ref = new_node
    return head_ref
 
# function for Sorted merge of two
# sorted doubly linked list
def merge(first, second):
     
    # If first list is empty
    if (first == None):
        return second
 
    # If second list is empty
    if (second == None):
        return first
 
    # Pick the smaller value and
    # adjust the links
    if (first.data < second.data) :
        first.next = merge(first.next,
                           second)
        first.next.prev = first
        first.prev = None
        return first
     
    else :
        second.next = merge(first,
                            second.next)
        second.next.prev = second
        second.prev = None
        return second
     
# function for Sorted merge of two sorted
# doubly circular linked list
def mergeUtil(head1, head2):
     
    # if 1st list is empty
    if (head1 == None):
        return head2
 
    # if 2nd list is empty
    if (head2 == None):
        return head1
 
    # get pointer to the node
    # which will be the last node
    # of the final list last_node
    if (head1.prev.data < head2.prev.data):
        last_node = head2.prev
    else:
        last_node = head1.prev
 
    # store None to the 'next' link of
    # the last nodes of the two lists
    head1.prev.next = None
    head2.prev.next = None
 
    # sorted merge of head1 and head2
    finalHead = merge(head1, head2)
 
    # 'prev' of 1st node pointing the last node
    # 'next' of last node pointing to 1st node
    finalHead.prev = last_node
    last_node.next = finalHead
 
    return finalHead
 
# function to print the list
def printList(head):
    temp = head
 
    while (temp.next != head):
        print(temp.data, end = " ")
        temp = temp.next
     
    print(temp.data, end = " ")
 
# Driver Code
if __name__=='__main__':
    head1 = None
    head2 = None
 
    # list 1:
    head1 = insert(head1, 8)
    head1 = insert(head1, 5)
    head1 = insert(head1, 3)
    head1 = insert(head1, 1)
 
    # list 2:
    head2 = insert(head2, 11)
    head2 = insert(head2, 9)
    head2 = insert(head2, 7)
    head2 = insert(head2, 2)
 
    newHead = mergeUtil(head1, head2)
 
    print("Final Sorted List: ", end = "")
    printList(newHead)
 
# This code is contributed by Srathore

C#




// C# implementation for Sorted merge of two
// sorted doubly circular linked list
using System;
 
class GFG
{
     
public class Node
{
    public int data;
    public Node next, prev;
};
 
// A utility function to insert a new node at the
// beginning of doubly circular linked list
static Node insert(Node head_ref, int data)
{
    // allocate space
    Node new_node = new Node();
 
    // put in the data
    new_node.data = data;
 
    // if list is empty
    if (head_ref == null)
    {
        new_node.next = new_node;
        new_node.prev = new_node;
    }
 
    else
    {
 
        // pointer points to last Node
        Node last = (head_ref).prev;
 
        // setting up previous and next of new node
        new_node.next = head_ref;
        new_node.prev = last;
 
        // update next and previous pointers of head_ref
        // and last.
        last.next = (head_ref).prev = new_node;
    }
 
    // update head_ref pointer
    head_ref = new_node;
    return head_ref;
}
 
// function for Sorted merge of two
// sorted doubly linked list
static Node merge(Node first, Node second)
{
    // If first list is empty
    if (first == null)
        return second;
 
    // If second list is empty
    if (second == null)
        return first;
 
    // Pick the smaller value and adjust
    // the links
    if (first.data < second.data)
    {
        first.next = merge(first.next, second);
        first.next.prev = first;
        first.prev = null;
        return first;
    }
    else
    {
        second.next = merge(first, second.next);
        second.next.prev = second;
        second.prev = null;
        return second;
    }
}
 
// function for Sorted merge of two sorted
// doubly circular linked list
static Node mergeUtil(Node head1, Node head2)
{
    // if 1st list is empty
    if (head1 == null)
        return head2;
 
    // if 2nd list is empty
    if (head2 == null)
        return head1;
 
    // get pointer to the node which will be the
    // last node of the final list
    Node last_node;
    if (head1.prev.data < head2.prev.data)
        last_node = head2.prev;
    else
        last_node = head1.prev;
 
    // store null to the 'next' link of the last nodes
    // of the two lists
    head1.prev.next = head2.prev.next = null;
 
    // sorted merge of head1 and head2
    Node finalHead = merge(head1, head2);
 
    // 'prev' of 1st node pointing the last node
    // 'next' of last node pointing to 1st node
    finalHead.prev = last_node;
    last_node.next = finalHead;
 
    return finalHead;
}
 
// function to print the list
static void printList(Node head)
{
    Node temp = head;
 
    while (temp.next != head)
    {
        Console.Write(temp.data + " ");
        temp = temp.next;
    }
    Console.Write(temp.data + " ");
}
 
// Driver code
public static void Main()
{
    Node head1 = null, head2 = null;
 
    // list 1:
    head1 = insert(head1, 8);
    head1 = insert(head1, 5);
    head1 = insert(head1, 3);
    head1 = insert(head1, 1);
 
    // list 2:
    head2 = insert(head2, 11);
    head2 = insert(head2, 9);
    head2 = insert(head2, 7);
    head2 = insert(head2, 2);
 
    Node newHead = mergeUtil(head1, head2);
 
    Console.Write( "Final Sorted List: ");
    printList(newHead);
}
}
 
// This code is contributed by Princi Singh

Javascript




<script>
// javascript implementation for Sorted merge of two
// sorted doubly circular linked list    
class Node
{
constructor()
{
         this.data = 0;
         this.next= this.prev = null;
}
}
 
    // A utility function to insert a new node at the
    // beginning of doubly circular linked list
    function insert( head_ref, data)
    {
     
        // allocate space
         new_node = new Node();
 
        // put in the data
        new_node.data = data;
 
        // if list is empty
        if (head_ref == null)
        {
            new_node.next = new_node;
            new_node.prev = new_node;
        }
 
        else
        {
 
            // pointer points to last Node
             last = (head_ref).prev;
 
            // setting up previous and next of new node
            new_node.next = head_ref;
            new_node.prev = last;
 
            // update next and previous pointers of head_ref
            // and last.
            last.next = (head_ref).prev = new_node;
        }
 
        // update head_ref pointer
        head_ref = new_node;
        return head_ref;
    }
 
    // function for Sorted merge of two
    // sorted doubly linked list
    function merge( first,  second)
    {
     
        // If first list is empty
        if (first == null)
            return second;
 
        // If second list is empty
        if (second == null)
            return first;
 
        // Pick the smaller value and adjust
        // the links
        if (first.data < second.data) {
            first.next = merge(first.next, second);
            first.next.prev = first;
            first.prev = null;
            return first;
        } else {
            second.next = merge(first, second.next);
            second.next.prev = second;
            second.prev = null;
            return second;
        }
    }
 
    // function for Sorted merge of two sorted
    // doubly circular linked list
    function mergeUtil( head1,  head2)
    {
     
        // if 1st list is empty
        if (head1 == null)
            return head2;
 
        // if 2nd list is empty
        if (head2 == null)
            return head1;
 
        // get pointer to the node which will be the
        // last node of the final list
         last_node = null;
        if (head1.prev.data < head2.prev.data)
            last_node = head2.prev;
        else
            last_node = head1.prev;
 
        // store null to the 'next' link of the last nodes
        // of the two lists
        head1.prev.next = head2.prev.next = null;
 
        // sorted merge of head1 and head2
         finalHead = merge(head1, head2);
 
        // 'prev' of 1st node pointing the last node
        // 'next' of last node pointing to 1st node
        finalHead.prev = last_node;
        last_node.next = finalHead;
 
        return finalHead;
    }
 
    // function to prvar the list
    function printList( head) {
         temp = head;
 
        while (temp.next != head) {
            document.write(temp.data + " ");
            temp = temp.next;
        }
        document.write(temp.data + " ");
    }
 
    // Driver code
     
         head1 = null, head2 = null;
 
        // list 1:
        head1 = insert(head1, 8);
        head1 = insert(head1, 5);
        head1 = insert(head1, 3);
        head1 = insert(head1, 1);
 
        // list 2:
        head2 = insert(head2, 11);
        head2 = insert(head2, 9);
        head2 = insert(head2, 7);
        head2 = insert(head2, 2);
 
         newHead = mergeUtil(head1, head2);
 
        document.write("Final Sorted List: ");
        printList(newHead);
 
// This code is contributed by umadevi9616
</script>

Output: 
 

Final Sorted List: 1 2 3 5 7 8 9 11

Time Complexity: O(n1 + n2).
 




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