Sorted insert for circular linked list
Difficulty Level: Rookie
Write a C function to insert a new value in a sorted Circular Linked List (CLL). For example, if the input CLL is following.
Algorithm:
Allocate memory for the newly inserted node and put data in the newly allocated node. Let the pointer to the new node be new_node. After memory allocation, following are the three cases that need to be handled.
1) Linked List is empty: a) since new_node is the only node in CLL, make a self loop. new_node->next = new_node; b) change the head pointer to point to new node. *head_ref = new_node; 2) New node is to be inserted just before the head node: (a) Find out the last node using a loop. while(current->next != *head_ref) current = current->next; (b) Change the next of last node. current->next = new_node; (c) Change next of new node to point to head. new_node->next = *head_ref; (d) change the head pointer to point to new node. *head_ref = new_node; 3) New node is to be inserted somewhere after the head: (a) Locate the node after which new node is to be inserted. while ( current->next!= *head_ref && current->next->data < new_node->data) { current = current->next; } (b) Make next of new_node as next of the located pointer new_node->next = current->next; (c) Change the next of the located pointer current->next = new_node;
C++
// C++ program for sorted insert // in circular linked list #include <bits/stdc++.h> using namespace std; /* structure for a node */ class Node { public : int data; Node *next; }; /* function to insert a new_node in a list in sorted way. Note that this function expects a pointer to head node as this can modify the head of the input linked list */ void sortedInsert(Node** head_ref, Node* new_node) { Node* current = *head_ref; // Case 1 of the above algo if (current == NULL) { new_node->next = new_node; *head_ref = new_node; } // Case 2 of the above algo else if (current->data >= new_node->data) { /* If value is smaller than head's value then we need to change next of last node */ while (current->next != *head_ref) current = current->next; current->next = new_node; new_node->next = *head_ref; *head_ref = new_node; } // Case 3 of the above algo else { /* Locate the node before the point of insertion */ while (current->next!= *head_ref && current->next->data < new_node->data) current = current->next; new_node->next = current->next; current->next = new_node; } } /* Function to print nodes in a given linked list */ void printList(Node *start) { Node *temp; if (start != NULL) { temp = start; do { cout<<temp->data<< " " ; temp = temp->next; } while (temp != start); } } /* Driver code */ int main() { int arr[] = {12, 56, 2, 11, 1, 90}; int list_size, i; /* start with empty linked list */ Node *start = NULL; Node *temp; /* Create linked list from the array arr[]. Created linked list will be 1->2->11->12->56->90 */ for (i = 0; i< 6; i++) { temp = new Node(); temp->data = arr[i]; sortedInsert(&start, temp); } printList(start); return 0; } // This code is contributed by rathbhupendra. |
C
#include<stdio.h> #include<stdlib.h> /* structure for a node */ struct Node { int data; struct Node *next; }; /* function to insert a new_node in a list in sorted way. Note that this function expects a pointer to head node as this can modify the head of the input linked list */ void sortedInsert( struct Node** head_ref, struct Node* new_node) { struct Node* current = *head_ref; // Case 1 of the above algo if (current == NULL) { new_node->next = new_node; *head_ref = new_node; } // Case 2 of the above algo else if (current->data >= new_node->data) { /* If value is smaller than head's value then we need to change next of last node */ while (current->next != *head_ref) current = current->next; current->next = new_node; new_node->next = *head_ref; *head_ref = new_node; } // Case 3 of the above algo else { /* Locate the node before the point of insertion */ while (current->next!= *head_ref && current->next->data < new_node->data) current = current->next; new_node->next = current->next; current->next = new_node; } } /* Function to print nodes in a given linked list */ void printList( struct Node *start) { struct Node *temp; if (start != NULL) { temp = start; printf ( "\n" ); do { printf ( "%d " , temp->data); temp = temp->next; } while (temp != start); } } /* Driver program to test above functions */ int main() { int arr[] = {12, 56, 2, 11, 1, 90}; int list_size, i; /* start with empty linked list */ struct Node *start = NULL; struct Node *temp; /* Create linked list from the array arr[]. Created linked list will be 1->2->11->12->56->90 */ for (i = 0; i< 6; i++) { temp = ( struct Node *) malloc ( sizeof ( struct Node)); temp->data = arr[i]; sortedInsert(&start, temp); } printList(start); return 0; } |
Java
// Java program for sorted insert in circular linked list class Node { int data; Node next; Node( int d) { data = d; next = null ; } } class LinkedList { Node head; // Constructor LinkedList() { head = null ; } /* function to insert a new_node in a list in sorted way. Note that this function expects a pointer to head node as this can modify the head of the input linked list */ void sortedInsert(Node new_node) { Node current = head; // Case 1 of the above algo if (current == null ) { new_node.next = new_node; head = new_node; } // Case 2 of the above algo else if (current.data >= new_node.data) { /* If value is smaller than head's value then we need to change next of last node */ while (current.next != head) current = current.next; current.next = new_node; new_node.next = head; head = new_node; } // Case 3 of the above algo else { /* Locate the node before the point of insertion */ while (current.next != head && current.next.data < new_node.data) current = current.next; new_node.next = current.next; current.next = new_node; } } // Utility method to print a linked list void printList() { if (head != null ) { Node temp = head; do { System.out.print(temp.data + " " ); temp = temp.next; } while (temp != head); } } // Driver code to test above public static void main(String[] args) { LinkedList list = new LinkedList(); // Creating the linkedlist int arr[] = new int [] { 12 , 56 , 2 , 11 , 1 , 90 }; /* start with empty linked list */ Node temp = null ; /* Create linked list from the array arr[]. Created linked list will be 1->2->11->12->56->90*/ for ( int i = 0 ; i < 6 ; i++) { temp = new Node(arr[i]); list.sortedInsert(temp); } list.printList(); } } // This code has been contributed by Mayank Jaiswal |
Python3
# Node class class Node: # Constructor to initialize the node object def __init__( self , data): self .data = data self . next = None class LinkedList: # Function to initialize head def __init__( self ): self .head = None # Function to insert a new node at the beginning def push( self , new_data): new_node = Node(new_data) new_node. next = self .head self .head = new_node # Utility function to print the linked LinkedList def printList( self ): temp = self .head print (temp.data,end = ' ' ) temp = temp. next while (temp ! = self .head): print (temp.data,end = ' ' ) temp = temp. next """ function to insert a new_node in a list in sorted way. Note that this function expects a pointer to head node as this can modify the head of the input linked list """ def sortedInsert( self , new_node): current = self .head # Case 1 of the above algo if current is None : new_node. next = new_node self .head = new_node # Case 2 of the above algo elif (current.data > = new_node.data): # If value is smaller than head's value then we # need to change next of last node while current. next ! = self .head : current = current. next current. next = new_node new_node. next = self .head self .head = new_node # Case 3 of the above algo else : # Locate the node before the point of insertion while (current. next ! = self .head and current. next .data < new_node.data): current = current. next new_node. next = current. next current. next = new_node # Driver program to test the above function #llist = LinkedList() arr = [ 12 , 56 , 2 , 11 , 1 , 90 ] list_size = len (arr) # start with empty linked list start = LinkedList() # Create linked list from the array arr[] # Created linked list will be 1->2->11->12->56->90 for i in range (list_size): temp = Node(arr[i]) start.sortedInsert(temp) start.printList() # This code is contributed by Nikhil Kumar Singh(nickzuck_007) |
C#
// C# program for sorted insert // in circular linked list using System; class LinkedList { public class Node { public int data; public Node next; public Node( int d) { data = d; next = null ; } } Node head; // Constructor LinkedList() { head = null ; } /* function to insert a new_node in a list in sorted way. Note that this function expects a pointer to head node as this can modify the head of the input linked list */ void sortedInsert(Node new_node) { Node current = head; // Case 1 of the above algo if (current == null ) { new_node.next = new_node; head = new_node; } // Case 2 of the above algo else if (current.data >= new_node.data) { /* If value is smaller than head's value then we need to change next of last node */ while (current.next != head) current = current.next; current.next = new_node; new_node.next = head; head = new_node; } // Case 3 of the above algo else { /* Locate the node before the point of insertion */ while (current.next != head && current.next.data < new_node.data) current = current.next; new_node.next = current.next; current.next = new_node; } } // Utility method to print a linked list void printList() { if (head != null ) { Node temp = head; do { Console.Write(temp.data + " " ); temp = temp.next; } while (temp != head); } } // Driver code public static void Main(String []args) { LinkedList list = new LinkedList(); // Creating the linkedlist int []arr = {12, 56, 2, 11, 1, 90}; /* start with empty linked list */ Node temp = null ; /* Create linked list from the array arr[]. Created linked list will be 1->2->11->12->56->90*/ for ( int i = 0; i < 6; i++) { temp = new Node(arr[i]); list.sortedInsert(temp); } list.printList(); } } // This code has been contributed // by Arnab Kundu |
Javascript
<script> // javascript program for sorted insert in circular linked list class Node { constructor(val) { this .data = val; this .next = null ; } } var head = null ; /* * function to insert a new_node in a list in sorted way. * Note that this function expects a pointer to head node * as this can modify the head of the * input linked list */ function sortedInsert(new_node) { var current = head; // Case 1 of the above algo if (current == null ) { new_node.next = new_node; head = new_node; } // Case 2 of the above algo else if (current.data >= new_node.data) { /* * If value is smaller than head's value then we * need to change next of last node */ while (current.next != head) current = current.next; current.next = new_node; new_node.next = head; head = new_node; } // Case 3 of the above algo else { /* Locate the node before the point of insertion */ while (current.next != head && current.next.data < new_node.data) current = current.next; new_node.next = current.next; current.next = new_node; } } // Utility method to print a linked list function printList() { if (head != null ) { var temp = head; do { document.write(temp.data + " " ); temp = temp.next; } while (temp != head); } } // Driver code to test above // Creating the linkedlist var arr = [ 12, 56, 2, 11, 1, 90 ]; /* start with empty linked list */ var temp = null ; /* * Create linked list from the array arr. * Created linked list will be * 1->2->11->12->56->90 */ for (i = 0; i < 6; i++) { temp = new Node(arr[i]); sortedInsert(temp); } printList(); // This code contributed by umadevi9616 </script> |
Output:
1 2 11 12 56 90
Time Complexity: O(n)
Here n is the number of nodes in the given linked list.
Auxiliary Space: O(1).
As constant extra space is used.
Case 2 of the above algorithm/code can be optimized. To implement the suggested change we need to modify case 2 to follow.
C
// Case 2 of the above algo else if (current->data >= new_node->data) { // swap the data part of head node and new node // assuming that we have a function swap(int *, int *) swap(&(current->data), &(new_node->data)); new_node->next = (*head_ref)->next; (*head_ref)->next = new_node; } |
C++
// Case 2 of the above algo else if (current->data >= new_node->data) { // swap the data part of head node and new node // assuming that we have a function swap(int *, int *) swap(&(current->data), &(new_node->data)); new_node->next = (*head_ref)->next; (*head_ref)->next = new_node; } // this code is contributed by devendra salunke |
Java
// Case 2 of the above algo else if (current.data >= new_node.data) { // swap the data part of head node and new node // assuming that we have a function swap(int *, int *) Node tmp = current.data; current.data = new_node.data; new_node.data = tmp; new_node.next = (head_ref).next; (head_ref).next = new_node; } // This code is contributed by pratham76. |
Python3
# Case 2 of the above algo elif (current.data > = new_node.data): # swap the data part of head node and new node # assuming that we have a function swap(int *, int *) tmp = current.data; current.data = new_node.data; new_node.data = tmp; new_node. next = (head_ref). next ; (head_ref). next = new_node; # This code is contributed by _saurabh_jaiswal |
C#
// Case 2 of the above algo else if (current.data >= new_node.data) { // swap the data part of head node and new node // assuming that we have a function swap(int *, int *) Node tmp = current.data; current.data = new_node.data; new_node.data = tmp; new_node.next = (head_ref).next; (head_ref).next = new_node; } // This code is contributed by rutvik_56 |
Javascript
<script> // Case 2 of the above algo else if (current.data >= new_node.data) { // swap the data part of head node and new node // assuming that we have a function swap(int *, int *) let tmp = current.data; current.data = new_node.data; new_node.data = tmp; new_node.next = (head_ref).next; (head_ref).next = new_node; } // This code is contributed by avanitrachhadiya2155 </script> |
In the previous approach the time complexity of case 2 was O(n) which is reduced to O(1) using this approach.
The auxiliary space used here is O(1) which is same as previous approach.
Please write comments if you find the above code/algorithm incorrect, or find other ways to solve the same problem.