Given an XOR linked list, the task is to find the middle node of the given XOR linked list.
Examples:
Input: 4 –> 7 –> 5
Output: 7
Explanation:
The middle node of the given XOR list is 7.Input: 4 –> 7 –> 5 –> 1
Output: 7 5
Explanation:
The two middle nodes of the XOR linked list with even number of nodes are 7 and 5.
Approach: Follow the steps below to solve the problem:
- Traverse to (Length / 2)th node of the Linked List.
- If the number of nodes is found to be odd, then print (Length + 1) / 2 th node as the only middle node.
- If the number of nodes is found to be even, then print both Length / 2 th node and (Length / 2) + 1 th node as the middle nodes.
Below is the implementation of the above approach:
C++
// C++ program to implement // the above approach #include <bits/stdc++.h> #include <inttypes.h> using namespace std; // Structure of a node // in XOR linked list struct Node { // Stores data value // of a node int data; // Stores XOR of previous // pointer and next pointer struct Node* nxp; }; // Function to find the XOR of two nodes struct Node* XOR( struct Node* a, struct Node* b) { return ( struct Node*)(( uintptr_t )(a) ^ ( uintptr_t )(b)); } // Function to insert a node with // given value at given position struct Node* insert( struct Node** head, int value) { // If XOR linked list is empty if (*head == NULL) { // Initialize a new Node struct Node* node = new Node; // Stores data value in // the node node->data = value; // Stores XOR of previous // and next pointer node->nxp = XOR(NULL, NULL); // Update pointer of head node *head = node; } // If the XOR linked list // is not empty else { // Stores the address // of current node struct Node* curr = *head; // Stores the address // of previous node struct Node* prev = NULL; // Initialize a new Node struct Node* node = new Node(); // Update curr node address curr->nxp = XOR(node, XOR(NULL, curr->nxp)); // Update new node address node->nxp = XOR(NULL, curr); // Update head *head = node; // Update data value of // current node node->data = value; } return *head; } // Function to print the middle node int printMiddle( struct Node** head, int len) { int count = 0; // Stores XOR pointer // in current node struct Node* curr = *head; // Stores XOR pointer of // in previous Node struct Node* prev = NULL; // Stores XOR pointer of // in next node struct Node* next; int middle = ( int )len / 2; // Traverse XOR linked list while (count != middle) { // Forward traversal next = XOR(prev, curr->nxp); // Update prev prev = curr; // Update curr curr = next; count++; } // If the length of the // linked list is odd if (len & 1) { cout << curr->data << " " ; } // If the length of the // linked list is even else { cout << prev->data << " " << curr->data << " " ; } } // Driver Code int main() { /* Create following XOR Linked List head --> 4 –> 7 –> 5 */ struct Node* head = NULL; insert(&head, 4); insert(&head, 7); insert(&head, 5); printMiddle(&head, 3); return (0); } |
C
// C program to implement // the above approach #include <inttypes.h> #include <stdio.h> #include <stdlib.h> // Structure of a node // in XOR linked list struct Node { // Stores data value // of a node int data; // Stores XOR of previous // pointer and next pointer struct Node* nxp; }; // Function to find the XOR of two nodes struct Node* XOR( struct Node* a, struct Node* b) { return ( struct Node*)(( uintptr_t )(a) ^ ( uintptr_t )(b)); } // Function to insert a node with // given value at given position struct Node* insert( struct Node** head, int value) { // If XOR linked list is empty if (*head == NULL) { // Initialize a new Node struct Node* node = ( struct Node*) malloc ( sizeof ( struct Node)); // Stores data value in // the node node->data = value; // Stores XOR of previous // and next pointer node->nxp = XOR(NULL, NULL); // Update pointer of head node *head = node; } // If the XOR linked list // is not empty else { // Stores the address // of current node struct Node* curr = *head; // Stores the address // of previous node struct Node* prev = NULL; // Initialize a new Node struct Node* node = ( struct Node*) malloc ( sizeof ( struct Node)); // Update curr node address curr->nxp = XOR( node, XOR(NULL, curr->nxp)); // Update new node address node->nxp = XOR(NULL, curr); // Update head *head = node; // Update data value of // current node node->data = value; } return *head; } // Function to print the middle node int printMiddle( struct Node** head, int len) { int count = 0; // Stores XOR pointer // in current node struct Node* curr = *head; // Stores XOR pointer of // in previous Node struct Node* prev = NULL; // Stores XOR pointer of // in next node struct Node* next; int middle = ( int )len / 2; // Traverse XOR linked list while (count != middle) { // Forward traversal next = XOR(prev, curr->nxp); // Update prev prev = curr; // Update curr curr = next; count++; } // If the length of the // linked list is odd if (len & 1) { printf ( "%d" , curr->data); } // If the length of the // linked list is even else { printf ( "%d %d" , prev->data, curr->data); } } // Driver Code int main() { /* Create following XOR Linked List head --> 4 –> 7 –> 5 */ struct Node* head = NULL; insert(&head, 4); insert(&head, 7); insert(&head, 5); printMiddle(&head, 3); return (0); } |
Python3
# C program to implement # the above approach import ctypes # Structure of a node in XOR linked list class Node: def __init__( self , value): self .value = value self .npx = 0 # create linked list class class XorLinkedList: # constructor def __init__( self ): self .head = None self .tail = None self .__nodes = [] # Function to insert a node with given value at given position def insert( self , value): # Initialize a new Node node = Node(value) # Check If XOR linked list is empty if self .head is None : # Update pointer of head node self .head = node # Update pointer of tail node self .tail = node else : # Update curr node address self .head.npx = id (node) ^ self .head.npx # Update new node address node.npx = id ( self .head) # Update head self .head = node # push node self .__nodes.append(node) # method to get length of linked list def Length( self ): if not self .isEmpty(): prev_id = 0 node = self .head next_id = 1 count = 1 while next_id: next_id = prev_id ^ node.npx if next_id: prev_id = id (node) node = self .__type_cast(next_id) count + = 1 else : return count else : return 0 # Function to print elements of the XOR Linked List def printMiddle( self , length): if self .head ! = None : prev_id = 0 node = self .head next_id = 1 # Traverse XOR linked list middle = length / / 2 count = 0 prev = None while count ! = middle: count = count + 1 # Forward traversal next_id = prev_id ^ node.npx if next_id: # Update prev prev_id = id (node) # Update curr prev = node node = self .__type_cast(next_id) else : return if length % 2 ! = 0 : print (node.value, end = ' ' ) else : print (prev.value, node.value, end = ' ' ) # method to check if the linked list is empty or not def isEmpty( self ): if self .head is None : return True return False # method to return a new instance of type def __type_cast( self , id ): return ctypes.cast( id , ctypes.py_object).value # Create following XOR Linked List # head-->40<-->30<-->20<-->10 head = XorLinkedList() head.insert( 4 ) head.insert( 7 ) head.insert( 5 ) # Reverse the XOR Linked List to give # head-->10<-->20<-->30<-->40 length = head.Length() head.printMiddle(length) # This code is contributed by Nidhi goel. |
Javascript
// JavaScript program to implement the above approach class Node { constructor(value) { this .value = value; this .npx = 0; } } // create linked list class class XorLinkedList { // constructor constructor() { this .head = null ; this .tail = null ; this .__nodes = []; } // Function to insert a node with given value at given position insert(value) { // Initialize a new Node const node = new Node(value); // Check If XOR linked list is empty if ( this .head === null ) { // Update pointer of head node this .head = node; // Update pointer of tail node this .tail = node; } else { // Update curr node address this .head.npx = this .__getId(node) ^ this .head.npx; // Update new node address node.npx = this .__getId( this .head); // Update head this .head = node; } // push node this .__nodes.push(node); } // method to get length of linked list length() { if (! this .isEmpty()) { let prevId = 0; let node = this .head; let nextId = 1; let count = 1; while (nextId) { nextId = prevId ^ node.npx; if (nextId) { prevId = this .__getId(node); node = this .__typeCast(nextId); count += 1; } else { return count; } } } else { return 0; } } // Function to print elements of the XOR Linked List printMiddle(length) { if ( this .head !== null ) { let prevId = 0; let node = this .head; let nextId = 1; // Traverse XOR linked list const middle = Math.floor(length / 2); let count = 0; let prev = null ; while (count !== middle) { count = count + 1; // Forward traversal nextId = prevId ^ node.npx; if (nextId) { // Update prev prevId = this .__getId(node); // Update curr prev = node; node = this .__typeCast(nextId); } else { return ; } } if (length % 2 !== 0) { console.log(node.value); } else { console.log(prev.value, node.value); } } } // method to check if the linked list is empty or not isEmpty() { if ( this .head === null ) { return true ; } return false ; } // method to return a new instance of type __typeCast(id) { return ctypes.cast(id, ctypes.py_object).value; } // method to get the unique ID of an object __getId(obj) { return obj && obj.__unique_id__; } } // Create following XOR Linked List // head-->40<-->30<-->20<-->10 const head = new XorLinkedList(); head.insert(4); head.insert(5); head.insert(7); // Reverse the XOR Linked List to give // head-->10<-->20<-->30<-->40 const length = head.length(); head.printMiddle(length); //this is generated by chetanbargal |
Output:
7
Time Complexity: O(N)
Auxiliary Space: O(1)
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