25. Reverse Nodes in k-Group

Reverse by group

class Solution {
    public ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || k == 1) return head;

        ListNode dummy = new ListNode(0);
        dummy.next = head;

        ListNode pre = dummy, end = dummy;

        while (true) {
            // check if there are k nodes
            for (int i = 0; i < k && end != null; i++) {
                end = end.next;
            }
            if (end == null) break;

            ListNode start = pre.next;
            ListNode next = end.next;
            end.next = null; // cut the list

            pre.next = reverse(start); // reverse the current part of the list

            start.next = next; // connect to the remaining list
            pre = start;
            end = pre;
        }

        return dummy.next;
    }

    // standard method for reverse a list
    private ListNode reverse(ListNode head) {
        ListNode prev = null;
        while (head != null) {
            ListNode nextTemp = head.next;
            head.next = prev;
            prev = head;
            head = nextTemp;
        }
        return prev;
    }
}

Remarks:

1. TC: $O(n)$, SC: $O(1)$