Deletion in Doubly Circular Linked List
We have discussed the doubly circular linked list introduction and its insertion.
Let us formulate the problem statement to understand the deletion process. Given a ‘key’, delete the first occurrence of this key in the circular doubly linked list.
Algorithm:
Case 1: Empty List(start = NULL)
- If the list is empty, simply return it.
Case 2: The List initially contains some nodes, start points at the first node of the List
- If the list is not empty, then we define two pointers curr and prev_1 and initialize the pointer curr points to the first node of the list, and prev_1 = NULL.
- Traverse the list using the curr pointer to find the node to be deleted and before moving from curr to the next node, every time set prev_1 = curr.
- If the node is found, check if it is the only node in the list. If yes, set start = NULL and free the node pointing by curr.
- If the list has more than one node, check if it is the first node of the list. The condition to check this is (curr == start). If yes, then move prev_1 to the last node(prev_1 = start -> prev). After prev_1 reaches the last node, set start = start -> next and prev_1 -> next = start and start -> prev = prev_1. Free the node pointing by curr.
- If curr is not the first node, we check if it is the last node in the list. The condition to check this is (curr -> next == start). If yes, set prev_1 -> next = start and start -> prev = prev_1. Free the node pointing by curr.
- If the node to be deleted is neither the first node nor the last node, declare one more pointer temp and initialize the pointer temp points to the next of curr pointer (temp = curr->next). Now set, prev_1 -> next = temp and temp ->prev = prev_1. Free the node pointing by curr.
- If the given key(Say 4) matches with the first node of the list(Step 4):
- If the given key(Say 8) matches with the last node of the list(Step 5):
- If the given key(Say 6) matches with the middle node of the list(Step 6):
Implementation:
C++
// C++ program to delete a given key from// circular doubly linked list.#include <bits/stdc++.h>using namespace std; // Structure of a Nodestruct Node { int data; struct Node* next; struct Node* prev;}; // Function to insert node in the listvoid insert(struct Node** start, int value){ // If the list is empty, create a single node // circular and doubly list if (*start == NULL) { struct Node* new_node = new Node; new_node->data = value; new_node->next = new_node->prev = new_node; *start = new_node; return; } // If list is not empty /* Find last node */ Node* last = (*start)->prev; // Create Node dynamically struct Node* new_node = new Node; new_node->data = value; // Start is going to be next of new_node new_node->next = *start; // Make new node previous of start (*start)->prev = new_node; // Make last previous of new node new_node->prev = last; // Make new node next of old last last->next = new_node;} // Function to delete a given node from the listvoid deleteNode(struct Node** start, int key){ // If list is empty if (*start == NULL) return; // Find the required node // Declare two pointers and initialize them struct Node *curr = *start, *prev_1 = NULL; while (curr->data != key) { // If node is not present in the list if (curr->next == *start) { printf("\nList doesn't have node with value = %d", key); return; } prev_1 = curr; curr = curr->next; } // Check if node is the only node in list if (curr->next == *start && prev_1 == NULL) { (*start) = NULL; free(curr); return; } // If list has more than one node, // check if it is the first node if (curr == *start) { // Move prev_1 to last node prev_1 = (*start)->prev; // Move start ahead *start = (*start)->next; // Adjust the pointers of prev_1 and start node prev_1->next = *start; (*start)->prev = prev_1; free(curr); } // check if it is the last node else if (curr->next == *start) { // Adjust the pointers of prev_1 and start node prev_1->next = *start; (*start)->prev = prev_1; free(curr); } else { // create new pointer, points to next of curr node struct Node* temp = curr->next; // Adjust the pointers of prev_1 and temp node prev_1->next = temp; temp->prev = prev_1; free(curr); }} // Function to display list elementsvoid display(struct Node* start){ struct Node* temp = start; while (temp->next != start) { printf("%d ", temp->data); temp = temp->next; } printf("%d ", temp->data);} // Driver program to test above functionsint main(){ // Start with the empty list struct Node* start = NULL; // Created linked list will be 4->5->6->7->8 insert(&start, 4); insert(&start, 5); insert(&start, 6); insert(&start, 7); insert(&start, 8); printf("List Before Deletion: "); display(start); // Delete the node which is not present in list deleteNode(&start, 9); printf("\nList After Deletion: "); display(start); // Delete the first node deleteNode(&start, 4); printf("\nList After Deleting %d: ", 4); display(start); // Delete the last node deleteNode(&start, 8); printf("\nList After Deleting %d: ", 8); display(start); // Delete the middle node deleteNode(&start, 6); printf("\nList After Deleting %d: ", 6); display(start); return 0;} |
Java
// Java program to delete a given key from// circular doubly linked list.import java.util.*;import java.io.*; class GFG { // structure of a Node static class Node { int data; Node next; Node prev; }; // Function to insert node in the list static Node insert(Node start, int value) { // If the list is empty, create a single node // circular and doubly list if (start == null) { Node new_node = new Node(); new_node.data = value; new_node.next = new_node.prev = new_node; start = new_node; return start; } // If list is not empty // Find last node / Node last = (start).prev; // Create Node dynamically Node new_node = new Node(); new_node.data = value; // Start is going to be next of new_node new_node.next = start; // Make new node previous of start (start).prev = new_node; // Make last previous of new node new_node.prev = last; // Make new node next of old last last.next = new_node; return start; } // Function to delete a given node from the list static Node deleteNode(Node start, int key) { // If list is empty if (start == null) return null; // Find the required node // Declare two pointers and initialize them Node curr = start, prev_1 = null; while (curr.data != key) { // If node is not present in the list if (curr.next == start) { System.out.printf("\nList doesn't have node with value = %d", key); return start; } prev_1 = curr; curr = curr.next; } // Check if node is the only node in list if (curr.next == start && prev_1 == null) { (start) = null; return start; } // If list has more than one node, // check if it is the first node if (curr == start) { // Move prev_1 to last node prev_1 = (start).prev; // Move start ahead start = (start).next; // Adjust the pointers of prev_1 and start node prev_1.next = start; (start).prev = prev_1; } // check if it is the last node else if (curr.next == start) { // Adjust the pointers of prev_1 and start node prev_1.next = start; (start).prev = prev_1; } else { // create new pointer, points to next of curr node Node temp = curr.next; // Adjust the pointers of prev_1 and temp node prev_1.next = temp; temp.prev = prev_1; } return start; } // Function to display list elements static void display(Node start) { Node temp = start; while (temp.next != start) { System.out.printf("%d ", temp.data); temp = temp.next; } System.out.printf("%d ", temp.data); } // Driver program to test above functions public static void main(String args[]) { // Start with the empty list Node start = null; // Created linked list will be 4.5.6.7.8 start = insert(start, 4); start = insert(start, 5); start = insert(start, 6); start = insert(start, 7); start = insert(start, 8); System.out.printf("List Before Deletion: "); display(start); // Delete the node which is not present in list start = deleteNode(start, 9); System.out.printf("\nList After Deletion: "); display(start); // Delete the first node start = deleteNode(start, 4); System.out.printf("\nList After Deleting %d: ", 4); display(start); // Delete the last node start = deleteNode(start, 8); System.out.printf("\nList After Deleting %d: ", 8); display(start); // Delete the middle node start = deleteNode(start, 6); System.out.printf("\nList After Deleting %d: ", 6); display(start); }} // This code is contributed by Arnab Kundu |
Python3
# Python3 program to delete a given key from# circular doubly linked list. # structure of a node of linked list class Node: def __init__(self, data): self.data = data self.next = None self.prev = None def insert( start, value): # If the list is empty, create a single node # circular and doubly list if (start == None): new_node = Node(0) new_node.data = value new_node.next = new_node.prev = new_node start = new_node return start # If list is not empty # Find last node / last = (start).prev # Create Node dynamically new_node = Node(0) new_node.data = value # Start is going to be next of new_node new_node.next = start # Make new node previous of start (start).prev = new_node # Make last previous of new node new_node.prev = last # Make new node next of old last last.next = new_node return start # Function to delete a given node# from the listdef deleteNode(start, key): # If list is empty if (start == None): return None # Find the required node # Declare two pointers and initialize them curr = start prev_1 = None while (curr.data != key) : # If node is not present in the list if (curr.next == start) : print ("List doesn't have node", "with value = ", key) return start prev_1 = curr curr = curr.next # Check if node is the only node in list if (curr.next == start and prev_1 == None) : (start) = None return start # If list has more than one node, # check if it is the first node if (curr == start) : # Move prev_1 to last node prev_1 = (start).prev # Move start ahead start = (start).next # Adjust the pointers of prev_1 # and start node prev_1.next = start (start).prev = prev_1 # check if it is the last node elif (curr.next == start) : # Adjust the pointers of prev_1 # and start node prev_1.next = start (start).prev = prev_1 else : # create new pointer, # points to next of curr node temp = curr.next # Adjust the pointers of prev_1 # and temp node prev_1.next = temp temp.prev = prev_1 return start # Function to display list elementsdef display(start): temp = start while (temp.next != start) : print (temp.data, end = " ") temp = temp.next print (temp.data) # Driver Code if __name__=='__main__': # Start with the empty list start = None # Created linked list will be 4.5.6.7.8 start = insert(start, 4) start = insert(start, 5) start = insert(start, 6) start = insert(start, 7) start = insert(start, 8) print ("List Before Deletion: ") display(start) # Delete the node which is not present in list start = deleteNode(start, 9) print ("List After Deletion: ") display(start) # Delete the first node start = deleteNode(start, 4) print ("List After Deleting", 4) display(start) # Delete the last node start = deleteNode(start, 8) print ("List After Deleting ", 8) display(start) # Delete the middle node start = deleteNode(start, 6) print ("List After Deleting ", 6) display(start) # This code is contributed by Arnab Kundu |
C#
// C# program to delete a given key from// circular doubly linked list.using System; class GFG { // structure of a Node public class Node { public int data; public Node next; public Node prev; }; // Function to insert node in the list static Node insert(Node start, int value) { // If the list is empty, create a single node // circular and doubly list Node new_node = new Node(); if (start == null) { new_node.data = value; new_node.next = new_node.prev = new_node; start = new_node; return start; } // If list is not empty // Find last node / Node last = (start).prev; // Create Node dynamically new_node = new Node(); new_node.data = value; // Start is going to be next of new_node new_node.next = start; // Make new node previous of start (start).prev = new_node; // Make last previous of new node new_node.prev = last; // Make new node next of old last last.next = new_node; return start; } // Function to delete a given node from the list static Node deleteNode(Node start, int key) { // If list is empty if (start == null) return null; // Find the required node // Declare two pointers and initialize them Node curr = start, prev_1 = null; while (curr.data != key) { // If node is not present in the list if (curr.next == start) { Console.Write("\nList doesn't have node with value = {0}", key); return start; } prev_1 = curr; curr = curr.next; } // Check if node is the only node in list if (curr.next == start && prev_1 == null) { (start) = null; return start; } // If list has more than one node, // check if it is the first node if (curr == start) { // Move prev_1 to last node prev_1 = (start).prev; // Move start ahead start = (start).next; // Adjust the pointers of prev_1 and start node prev_1.next = start; (start).prev = prev_1; } // check if it is the last node else if (curr.next == start) { // Adjust the pointers of prev_1 and start node prev_1.next = start; (start).prev = prev_1; } else { // create new pointer, points to next of curr node Node temp = curr.next; // Adjust the pointers of prev_1 and temp node prev_1.next = temp; temp.prev = prev_1; } return start; } // Function to display list elements static void display(Node start) { Node temp = start; while (temp.next != start) { Console.Write("{0} ", temp.data); temp = temp.next; } Console.Write("{0} ", temp.data); } // Driver code public static void Main(String[] args) { // Start with the empty list Node start = null; // Created linked list will be 4.5.6.7.8 start = insert(start, 4); start = insert(start, 5); start = insert(start, 6); start = insert(start, 7); start = insert(start, 8); Console.Write("List Before Deletion: "); display(start); // Delete the node which is not present in list start = deleteNode(start, 9); Console.Write("\nList After Deletion: "); display(start); // Delete the first node start = deleteNode(start, 4); Console.Write("\nList After Deleting {0}: ", 4); display(start); // Delete the last node start = deleteNode(start, 8); Console.Write("\nList After Deleting {0}: ", 8); display(start); // Delete the middle node start = deleteNode(start, 6); Console.Write("\nList After Deleting {0}: ", 6); display(start); }} // This code has been contributed by 29AjayKumar |
Javascript
<script> // javascript program to delete a given key from // circular doubly linked list. // structure of a Node class Node { constructor() { this.data = 0; this.prev = null; this.next = null; } } // Function to insert node in the list function insert(start , value) { // If the list is empty, create a single node // circular and doubly list if (start == null) { var new_node = new Node(); new_node.data = value; new_node.next = new_node.prev = new_node; start = new_node; return start; } // If list is not empty // Find last node / var last = (start).prev; // Create Node dynamically var new_node = new Node(); new_node.data = value; // Start is going to be next of new_node new_node.next = start; // Make new node previous of start (start).prev = new_node; // Make last previous of new node new_node.prev = last; // Make new node next of old last last.next = new_node; return start; } // Function to delete a given node from the list function deleteNode(start , key) { // If list is empty if (start == null) return null; // Find the required node // Declare two pointers and initialize them var curr = start, prev_1 = null; while (curr.data != key) { // If node is not present in the list if (curr.next == start) { document.write("<br/>List doesn't have node with value = "+ key); return start; } prev_1 = curr; curr = curr.next; } // Check if node is the only node in list if (curr.next == start && prev_1 == null) { (start) = null; return start; } // If list has more than one node, // check if it is the first node if (curr == start) { // Move prev_1 to last node prev_1 = (start).prev; // Move start ahead start = (start).next; // Adjust the pointers of prev_1 and start node prev_1.next = start; (start).prev = prev_1; } // check if it is the last node else if (curr.next == start) { // Adjust the pointers of prev_1 and start node prev_1.next = start; (start).prev = prev_1; } else { // create new pointer, points to next of curr node var temp = curr.next; // Adjust the pointers of prev_1 and temp node prev_1.next = temp; temp.prev = prev_1; } return start; } // Function to display list elements function display(start) { var temp = start; while (temp.next != start) { document.write( temp.data+" "); temp = temp.next; } document.write( temp.data+" "); } // Driver program to test above functions // Start with the empty list var start = null; // Created linked list will be 4.5.6.7.8 start = insert(start, 4); start = insert(start, 5); start = insert(start, 6); start = insert(start, 7); start = insert(start, 8); document.write("List Before Deletion: "); display(start); // Delete the node which is not present in list start = deleteNode(start, 9); document.write("<br/>List After Deletion: "); display(start); // Delete the first node start = deleteNode(start, 4); document.write("<br/>List After Deleting 4: "); display(start); // Delete the last node start = deleteNode(start, 8); document.write("<br/>List After Deleting 8: "); display(start); // Delete the middle node start = deleteNode(start, 6); document.write("<br/>List After Deleting 6: "); display(start); // This code contributed by aashish1995 </script> |
Output
List Before Deletion: 4 5 6 7 8 List doesn't have node with value = 9 List After Deletion: 4 5 6 7 8 List After Deleting 4: 5 6 7 8 List After Deleting 8: 5 6 7 List After Deleting 6: 5 7
Time Complexity: O(n), as we are using a loop to traverse n times (for deletion and displaying the linked list). Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.
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