> For the complete documentation index, see [llms.txt](https://mayanktyagi3111.gitbook.io/interview-prep/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://mayanktyagi3111.gitbook.io/interview-prep/linked-list/reverse-nodes-in-k-group.md).

# Reverse Nodes in k-Group

Given a linked list, reverse the nodes of a linked list *k* at a time and return its modified list.

*k* is a positive integer and is less than or equal to the length of the linked list. If the number of nodes is not a multiple of *k* then left-out nodes in the end should remain as it is.

*

**Example:**

Given this linked list: `1->2->3->4->5`

For *k* = 2, you should return: `2->1->4->3->5`

For *k* = 3, you should return: `3->2->1->4->5`

**Note:**

* Only constant extra memory is allowed.
* You may not alter the values in the list's nodes, only nodes itself may be changed.

```java
class Solution {
    public ListNode reverse(ListNode head) {
        ListNode prev = null, current = head;
        while (current != null) {
            ListNode temp = current.next;
            current.next = prev;
            prev = current;
            current = temp;
        }
        return prev;
    }

    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || head.next == null || k == 1)
            return head;
        ListNode previous = null, current = head;
        int index = 1;
        while (current != null) {
            if (index == k) {
                ListNode next = current.next;
                current.next = null;
                if (previous == null) {
                    previous = head;
                    head = reverse(head);
                    previous.next = next;
                } else {
                    ListNode temp = previous.next;
                    previous.next = reverse(previous.next);
                    temp.next = next;
                    previous = temp;
                }
                current = next;
                index = 1;
                continue;
            }
            current = current.next;
            index++;
        }
        return head;
    }
}
```