XOR Linked List – Reversal of a List
Given a XOR linked list, the task is to reverse the XOR linked list.
Examples:
Input: 4 <–> 7 <–> 9 <–> 7
Output: 7 <–> 9 <–> 7 <–> 4
Explanation:
Reversing the linked list modifies the XOR linked list to 7 <–> 9 <–> 7 <–> 4.Input: 2 <-> 5 <-> 7 <-> 4
Output: 4 <-> 7 <-> 5 <-> 2
Explanation:
Reversing the linked list modifies the XOR linked list to 4 <-> 7 <-> 5 <-> 2.
Approach: XOR linked listconsists of a single pointer, which is the only pointer needed to traverse the XOR linked list in both directions. Therefore, the idea to solve this problem is only by making the last node of the XOR linked list its Head Node. Follow the steps below to solve the problem:
- Initialize a pointer variable, say curr, to pointto the current node being traversed.
- Store the current head pointer in the curr variable.
- If curr is equal to NULL, then return NULL.
- Otherwise, traverse upto the last node and make it the head of the XOR linked list.
Below is the implementation of the above approach:
C++
// C++ program for the above approach#include<bits/stdc++.h>using namespace std;// Structure of a node// in XOR linked liststruct Node{ // Stores data value // of a node int data; // Stores XOR of previous // pointer and next pointer Node* nxp;};// Function to find the XOR of two nodesNode* XOR(Node* a, Node* b){ return (Node*)((uintptr_t)(a) ^ (uintptr_t)(b));}// Function to insert a node with// given value at begining positionNode* insert(Node** head, int value){ // If XOR linked list is empty if (*head == NULL) { // Initialize a new Node 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 Node* curr = *head; // Stores the address // of previous node Node* prev = NULL; // Initialize a new Node 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 elements of// the XOR Linked Listvoid printList(Node** head){ // Stores XOR pointer // in current node Node* curr = *head; // Stores XOR pointer of // in previous Node Node* prev = NULL; // Stores XOR pointer of // in next node Node* next; // Traverse XOR linked list while (curr != NULL) { // Print current node cout<<curr->data<<" "; // Forward traversal next = XOR(prev, curr->nxp); // Update prev prev = curr; // Update curr curr = next; } cout << endl;}// Function to reverse the XOR linked listNode* reverse(Node** head){ // Stores XOR pointer // in current node Node* curr = *head; if (curr == NULL) return NULL; else { // Stores XOR pointer of // in previous Node Node* prev = NULL; // Stores XOR pointer of // in next node Node* next; while (XOR(prev, curr->nxp) != NULL) { // Forward traversal next = XOR(prev, curr->nxp); // Update prev prev = curr; // Update curr curr = next; } // Update the head pointer *head = curr; return *head; }}// Driver Codeint main(){ /* Create following XOR Linked List head-->40<-->30<-->20<-->10 */ Node* head = NULL; insert(&head, 10); insert(&head, 20); insert(&head, 30); insert(&head, 40); /* Reverse the XOR Linked List to give head-->10<-->20<-->30<-->40 */ cout << "XOR linked list: "; printList(&head); reverse(&head); cout << "Reversed XOR linked list: "; printList(&head); return 0;}// This code is contributed by rutvik_56. |
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C
// C program for the above approach#include <inttypes.h>#include <stdio.h>#include <stdlib.h>// Structure of a node// in XOR linked liststruct 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 nodesstruct 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 begining positionstruct 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 elements of// the XOR Linked Listvoid printList(struct Node** head){ // 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; // Traverse XOR linked list while (curr != NULL) { // Print current node printf("%d ", curr->data); // Forward traversal next = XOR(prev, curr->nxp); // Update prev prev = curr; // Update curr curr = next; } printf("\n");}// Function to reverse the XOR linked liststruct Node* reverse(struct Node** head){ // Stores XOR pointer // in current node struct Node* curr = *head; if (curr == NULL) return NULL; else { // Stores XOR pointer of // in previous Node struct Node* prev = NULL; // Stores XOR pointer of // in next node struct Node* next; while (XOR(prev, curr->nxp) != NULL) { // Forward traversal next = XOR(prev, curr->nxp); // Update prev prev = curr; // Update curr curr = next; } // Update the head pointer *head = curr; return *head; }}// Driver Codeint main(){ /* Create following XOR Linked List head-->40<-->30<-->20<-->10 */ struct Node* head = NULL; insert(&head, 10); insert(&head, 20); insert(&head, 30); insert(&head, 40); /* Reverse the XOR Linked List to give head-->10<-->20<-->30<-->40 */ printf("XOR linked list: "); printList(&head); reverse(&head); printf("Reversed XOR linked list: "); printList(&head); return (0);} |
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Output:
XOR linked list: 40 30 20 10 Reversed XOR linked list: 10 20 30 40
Time Complexity: O(N)
Auxiliary Space: O(1)
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