Sorted merge of two sorted doubly circular linked lists
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:
List 2:
Final list:
Approach: Following are the steps:
- If head1 == NULL, return head2.
- If head2 == NULL, return head1.
- 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.
- 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.
- Update last1.next = last2.next = NULL.
- 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.
- Update finalHead.prev = last_node and last_node.next = finalHead.
- 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; } |
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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 |
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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 |
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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 |
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Output:
Final Sorted List: 1 2 3 5 7 8 9 11
Time Complexity: O(n1 + n2).
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