LeetCode 25 Reverse Nodes in k-Group

小细节太多了，需要注意！！！

public ListNode reverseKGroup(ListNode head, int k) {
    if (head == null || head.next == null) {
        return head;
    }
    ListNode dummy = new ListNode(0);
    dummy.next = head;
    ListNode cur = dummy;
    
    while (cur != null) {
        cur = reverseK(cur, k);
    }
    return dummy.next;
}

// 1  ->  2  ->  3  ->  4  ->  5 ->  6  k = 2
//head         node
//             cur    next
// 2 -> 1   ->  3  ->  4  
private ListNode reverseK(ListNode head, int k) {
ListNode tail = head;
ListNode prev = head;
ListNode n1 = head.next; //翻转list中的第一个元素。
for (int i = 0; i < k; i++) {
tail = tail.next;
//这里要判断剩余部分不够k的数的时候。直接返回null，不进行任何操作。
if (tail == null) {
return null;
}
}

ListNode next = tail.next;
    //           1 ->  2  ->  3     |    ->  4  ->  5
    //   head  first        tail            next 

ListNode newHead = reverse(head.next, tail);
    //    head      3 -> 2  ->  1 ->   |   4  ->  5  -> null
    //            newH        first       next

prev.next = newHead;
n1.next = next;
//不要返回next。 把n1当作下一次调用的dummy node, 所以每次返回的都是dummy node
return n1;
}

private ListNode reverse(ListNode head, ListNode tail) {
    if (head == tail) {
        return head;
    }
    ListNode prev = null, cur = head;
    while (cur != tail) {
        ListNode next = cur.next;
        cur.next = prev;
        prev = cur;
        cur = next;
    }
    
    if (cur == null) {
        return prev;
    }
    cur.next = prev;
    return cur;
}