Given a linked list, write a function to reverse every k nodes (where k is an input to the function).
Example:
Input: 1->2->3->4->5->6->7->8->NULL, K = 3
Output: 3->2->1->6->5->4->8->7->NULL
Input: 1->2->3->4->5->6->7->8->NULL, K = 5
Output: 5->4->3->2->1->8->7->6->NULL
Algorithm: reverse(head, k)
- Reverse the first sub-list of size k. While reversing keep track of the next node and previous node. Let the pointer to the next node be next and pointer to the previous node be prev. See this post for reversing a linked list.
- head->next = reverse(next, k) ( Recursively call for rest of the list and link the two sub-lists )
- Return prev ( prev becomes the new head of the list (see the diagrams of an iterative method of this post )
Below is image shows how the reverse function works:
Below is the implementation of the above approach:
Javascript
<script> // Javascript program to reverse a // linked list in groups of // given size // Head of list var head; // Linked list Node class Node { constructor(val) { this.data = val; this.next = null; } } function reverse(head, k) { if (head == null) return null; var current = head; var next = null; var prev = null; var count = 0; // Reverse first k nodes of // linked list while (count < k && current != null) { next = current.next; current.next = prev; prev = current; current = next; count++; } /* next is now a pointer to (k+1)th node Recursively call for the list starting from current. And make rest of the list as next of first node. */ if (next != null) head.next = reverse(next, k); // prev is now head of input list return prev; } // Utility functions // Inserts a new Node at front // of the list. function push(new_data) { /* 1 & 2: Allocate the Node & Put in the data */ new_node = new Node(new_data); // 3. Make next of new Node as head new_node.next = head; // 4. Move the head to point to // new Node head = new_node; } // Function to print linked list function printList() { temp = head; while (temp != null) { document.write(temp.data + " "); temp = temp.next; } document.write("<br/>"); } // Driver code /* Create Linked List 1->2->3->4->5->6-> 7->8->8->9->null */push(9); push(8); push(7); push(6); push(5); push(4); push(3); push(2); push(1); document.write("Given Linked List<br/>"); printList(); head = reverse(head, 3); document.write("Reversed list<br/>"); printList(); // This code is contributed by gauravrajput1 </script> |
Output:
Given Linked List 1 2 3 4 5 6 7 8 9 Reversed list 3 2 1 6 5 4 9 8 7
Complexity Analysis:
- Time Complexity: O(n).
Traversal of list is done only once and it has ‘n’ elements. - Auxiliary Space: O(n/k).
For each Linked List of size n, n/k or (n/k)+1 calls will be made during the recursion.
Please refer complete article on Reverse a Linked List in groups of given size | Set 1 for more details!
Feeling lost in the world of random DSA topics, wasting time without progress? It’s time for a change! Join our DSA course, where we’ll guide you on an exciting journey to master DSA efficiently and on schedule.
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
Ready to dive in? Explore our Free Demo Content and join our DSA course, trusted by over 100,000 neveropen!
