25 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.

do it again and again

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || head.next == null) { // 5 return 
            return head;
        }
        ListNode curt = head;
        int count = 0;
        while (curt != null && count != k) {
            curt = curt.next;
            count++;
        }
        if (count == k) {
            curt = reverseKGroup(curt, k);// 3->4->5
            while (count-- > 0) {
                ListNode temp = head.next;
                head.next = curt;
                curt = head;
                head = temp;
            }
            head = curt;
        }
        // now curt points to 3 
        return head;
    }
}