解题思路

栈

用栈，我们把 k 个数压入栈中，然后弹出来的顺序就是翻转的！

这里要注意几个问题：

第一，剩下的链表个数够不够 k 个（因为不够 k 个不用翻转）；

第二，已经翻转的部分要与剩下链表连接起来。

 public ListNode reverseKGroup(ListNode head, int k) {
        Deque<ListNode> stack = new ArrayDeque<ListNode>();
        ListNode dummy = new ListNode(0);
        ListNode p = dummy;
        while (true) {
            int count = 0;
            ListNode tmp = head;
            while (tmp != null && count < k) {
                stack.add(tmp);
                tmp = tmp.next;
                count++;
            }
            if (count != k) {
                p.next = head;
                break;
            }
            while (!stack.isEmpty()){
                p.next = stack.pollLast();
                p = p.next;
            }
            p.next = tmp;
            head = tmp;
        }
        return dummy.next;
}

 public ListNode reverseKGroup(ListNode head, int k) {
        ListNode dummy = new ListNode(0);
        dummy.next = head;
        ListNode pre = dummy;
        ListNode tail = dummy;
        while (true) {
            int count = 0;
            while (tail != null && count != k) {
                count++;
                tail = tail.next;
            }
            if (tail == null) break;
            ListNode head1 = pre.next;
            while (pre.next != tail) {
                ListNode cur = pre.next;
                pre.next = cur.next;
                cur.next = tail.next;
                tail.next = cur;
            }
            pre = head1;
            tail = head1;
        }
        return dummy.next;
    }

递归

 public ListNode reverseKGroup(ListNode head, int k) {
        ListNode cur = head;
        int count = 0;
        while (cur != null && count != k) {
            cur = cur.next;
            count++;
        }
        if (count == k) {
            cur = reverseKGroup(cur, k);
            while (count != 0) {
                count--;
                ListNode tmp = head.next;
                head.next = cur;
                cur = head;
                head = tmp;
            }
            head = cur;
        }
        return head;
    }