Rotate Linked List block wise
Last Updated :
15 Jun, 2022
Improve
Given a Linked List of length n and block length k rotate in a circular manner towards right/left each block by a number d. If d is positive rotate towards right else rotate towards left.
Examples:
Input: 1->2->3->4->5->6->7->8->9->NULL,
k = 3
d = 1
Output: 3->1->2->6->4->5->9->7->8->NULL
Explanation: Here blocks of size 3 are
rotated towards right(as d is positive)
by 1.
Input: 1->2->3->4->5->6->7->8->9->10->
11->12->13->14->15->NULL,
k = 4
d = -1
Output: 2->3->4->1->6->7->8->5->10->11
->12->9->14->15->13->NULL
Explanation: Here, at the end of linked
list, remaining nodes are less than k, i.e.
only three nodes are left while k is 4.
Rotate those 3 nodes also by d.
Prerequisite: Rotate a linked list
The idea is if the absolute value of d is greater than the value of k, then rotate the link list by d % k times. If d is 0, no need to rotate the linked list at all.
C++
C
// C program to rotate a linked list block wise#include <stdio.h>#include <stdlib.h>/* Link list node */struct Node { int data; struct Node* next;};// Recursive function to rotate one blockstruct Node* rotateHelper(struct Node* blockHead, struct Node* blockTail, int d, struct Node** tail, int k){ if (d == 0) return blockHead; // Rotate Clockwise if (d > 0) { struct Node* temp = blockHead; for (int i = 1; temp->next->next && i < k - 1; i++) temp = temp->next; blockTail->next = blockHead; *tail = temp; return rotateHelper(blockTail, temp, d - 1, tail, k); } // Rotate anti-Clockwise if (d < 0) { blockTail->next = blockHead; *tail = blockHead; return rotateHelper(blockHead->next, blockHead, d + 1, tail, k); }}// Function to rotate the linked list block wisestruct Node* rotateByBlocks(struct Node* head, int k, int d){ // If length is 0 or 1 return head if (!head || !head->next) return head; // if degree of rotation is 0, return head if (d == 0) return head; struct Node* temp = head, *tail = NULL; // Traverse upto last element of this block int i; for (i = 1; temp->next && i < k; i++) temp = temp->next; // storing the first node of next block struct Node* nextBlock = temp->next; // If nodes of this block are less than k. // Rotate this block also if (i < k) head = rotateHelper(head, temp, d % k, &tail, i); else head = rotateHelper(head, temp, d % k, &tail, k); // Append the new head of next block to // the tail of this block tail->next = rotateByBlocks(nextBlock, k, d % k); // return head of updated Linked List return head;}/* UTILITY FUNCTIONS *//* Function to push a node */void push(struct Node** head_ref, int new_data){ struct Node* new_node = new Node; new_node->data = new_data; new_node->next = (*head_ref); (*head_ref) = new_node;}/* Function to print linked list */void printList(struct Node* node){ while (node != NULL) { printf("%d ", node->data); node = node->next; }}/* Driver program to test above function*/int main(){ /* Start with the empty list */ struct Node* head = NULL; // create a list 1->2->3->4->5-> // 6->7->8->9->NULL for (int i = 9; i > 0; i -= 1) push(&head, i); printf("Given linked list \n"); printList(head); // k is block size and d is number of // rotations in every block. int k = 3, d = 2; head = rotateByBlocks(head, k, d); printf("\nRotated by blocks Linked list \n"); printList(head); return (0);} |
Java
Python3
C#
Javascript
Output:
Given linked list 1 2 3 4 5 6 7 8 9 Rotated by blocks Linked list 2 3 1 5 6 4 8 9 7
Time complexity: O(n) since using a single loop to traverse a linked list to rotate
Auxiliary Space: O(n) for call stack
