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.

If the number of nodes is not a multiple of k then left-out nodes in the end should remain as it is.

You may not alter the values in the nodes, only nodes itself may be changed.

Only constant memory is allowed.

For 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

This problem is not a really "hard" one as long as you understand how to reverse lists, which I am always confused about. For that reason, I am posting a reverseList method for my reference.

        
ListNode reverseList(ListNode head)
{
    ListNode newhead = null;
    while (head != null)
    {
 ListNode tmp = head.next;
 head.next = newhead;
 newhead = head;
 head = tmp;
    }
    return new head;

}

1. We get the length of the list.
2. We divide the length by k, and get the number of "reverse" we need. Return boundary conditions.
3. We iterate to reverse the list k times. Remember to track the "tail" and "head" of each group to link them together.

public class ReverseList {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null)
            return null;
        int length = 0;
        ListNode curr = head;
        while (curr != null)
        {
            length++;
            curr = curr.next;
        }

        //in the case where k is greater than the length of the lists, return
        int multi = length / k;
        if (k <= 1 || multi == 0)
            return head;
        ListNode curtail = null;
        ListNode curhead = null;
        ListNode pretail = null;
        curr = head;
        for (int i = 0; i < multi; i++)
        {
            ListNode preNode = null;
            ListNode nextNode = null;
            for (int j = 0; j < k; j++)
            {
                nextNode = curr.next;
                curr.next = preNode;
                preNode = curr;
                if (j == 0) curtail = curr;
                if (j == k - 1) curhead = curr;
                curr = nextNode;
            }
            if (pretail != null)
                pretail.next = curhead;
            else
                head = curhead;
            pretail = curtail;
        }

        //if there are left-out nodes, current tail points to the next node

        // otherwise points to null
        curtail.next = curr;
        return head;
    }
}