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.

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

Solution: Use a variable i track k group.

public ListNode reverseKGroup(ListNode head, int k) {
    if(head == null || head.next == null) {
        return head;
    }
    ListNode dummy = new ListNode(-1);
    dummy.next = head;
    int i = 0;
    ListNode prev = dummy;
    while(head != null) {
        i++;
        ListNode tmp = head.next;
        if(i % k == 0) {
            prev = reverse(prev, head.next);
        }
        head = tmp;
    }
    return dummy.next;
}

public ListNode reverse(ListNode pre, ListNode end) {
    ListNode cur = pre.next, second = cur.next;
    while(cur.next != end) {
        cur.next = second.next;
        second.next = pre.next;
        pre.next = second;
        second = cur.next;
    }
    return cur;
}