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 <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 given 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 the middle nodeint 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 Codeint 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); } |
7
Time Complexity: O(N)
Auiliary Space: O(1)
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