Java Program For Reversing A Linked List In Groups Of Given Size – Set 1

Last Updated : 16 Dec, 2021

Given a linked list, write a function to reverse every k nodes (where k is an input to the function).Â

Example:Â

Input: 1->2->3->4->5->6->7->8->NULL, K = 3Â
Output: 3->2->1->6->5->4->8->7->NULLÂ
Input: 1->2->3->4->5->6->7->8->NULL, K = 5Â
Output: 5->4->3->2->1->8->7->6->NULLÂ

• 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:Â

Below is the implementation of the above approach:

Java

 `// Java program to reverse a  ` `// linked list in groups of ` `// given size ` `class` `LinkedList ` `{ ` `    ``// Head of list ` `    ``Node head;  ` ` `  `    ``// Linked list Node ` `    ``class` `Node  ` `    ``{ ` `        ``int` `data; ` `        ``Node next; ` `        ``Node(``int` `d) ` `        ``{ ` `            ``data = d; ` `            ``next = ``null``; ` `        ``} ` `    ``} ` ` `  `    ``Node reverse(Node head, ``int` `k) ` `    ``{ ` `        ``if``(head == ``null``) ` `          ``return` `null``; ` `        ``Node current = head; ` `        ``Node next = ``null``; ` `        ``Node prev = ``null``; ` ` `  `        ``int` `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 the input list ` `        ``return` `prev; ` `    ``} ` ` `  `    ``// Utility functions ` ` `  `    ``// Inserts a new Node at front ` `    ``// of the list. ` `    ``public` `void` `push(``int` `new_data) ` `    ``{ ` `        ``/* 1 & 2: Allocate the Node & ` `                  ``Put in the data*/` `        ``Node 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 ` `    ``void` `printList() ` `    ``{ ` `        ``Node temp = head; ` `        ``while` `(temp != ``null``)  ` `        ``{ ` `            ``System.out.print(temp.data + ``" "``); ` `            ``temp = temp.next; ` `        ``} ` `        ``System.out.println(); ` `    ``} ` ` `  `    ``// Driver code ` `    ``public` `static` `void` `main(String args[]) ` `    ``{ ` `        ``LinkedList llist = ``new` `LinkedList(); ` ` `  `        ``// Create Linked List is  ` `        ``// 1->2->3->4->5->6-> ` `        ``// 7->8->8->9->null ` `        ``llist.push(``9``); ` `        ``llist.push(``8``); ` `        ``llist.push(``7``); ` `        ``llist.push(``6``); ` `        ``llist.push(``5``); ` `        ``llist.push(``4``); ` `        ``llist.push(``3``); ` `        ``llist.push(``2``); ` `        ``llist.push(``1``); ` ` `  `        ``System.out.println(``"Given Linked List"``); ` `        ``llist.printList(); ` ` `  `        ``llist.head = llist.reverse(llist.head, ``3``); ` ` `  `        ``System.out.println(``"Reversed list"``); ` `        ``llist.printList(); ` `    ``} ` `} ` `// This code is contributed by Rajat Mishra  `

Output:Â

```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:Â

• Time Complexity: O(n).Â
Traversal of list is done only once and it has ‘n’ elements.
• Auxiliary Space: O(n/k).Â
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!

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