Reverse Nodes in k-Group

Problem: 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. 

Input: head = [1,2,3,4,5], k = 2
Output: [2,1,4,3,5]

Input: head = [1,2,3,4,5], k = 3
Output: [3,2,1,4,5]

Approach: Same as Reverse a Linked List in groups of given size except we need to check if the remaining length is sufficient or not.

Implementation in C#:

    public ListNode ReverseKGroup(ListNode head, int k)

    {

        if (!this.IsSuffLength(head, k))

        {

            return head;

        }

        ListNode prev = null, next = null, curr = head;

        int count = 0;

        while (count < k)

        {

            next = curr.next;

            curr.next = prev;

            prev = curr;

            curr = next;

            ++count;

        }

        if (next != null)

        {

            head.next = this.ReverseKGroup(next, k);

        }

        return prev;

    }

    private bool IsSuffLength(ListNode node, int k)

    {

        int i = 0;

        for (; node != null && i < k; ++i)

        {

            node = node.next;

        }

        if (i != k)

        {

            return false;

        }

        return true;

    }

Complexity: O(n)