递归解决k个一组的链表节点翻转问题

public class SwapKGroup {
    public static ListNode reverseKGroup(ListNode head, int k) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode tail = head;
        for (int i = 0; i < k; i++) {
            //剩余数量小于k的话，则不需要反转。
            if (tail == null) {
                return head;
            }
            tail = tail.next;
        }
        // 反转前 k 个元素
        ListNode newHead = reverse(head, tail);
        //下一轮的开始的地方就是tail
        head.next = reverseKGroup(tail, k);

        return newHead;
    }

    /*
    左闭右开区间
     */
    private static ListNode reverse(ListNode head, ListNode tail) {
        ListNode pre = null;
        ListNode next = null;
        while (head != tail) {
            next = head.next;
            head.next = pre;
            pre = head;
            head = next;
        }
        return pre;

    }

    public static void main(String[] args) {
        ListNode listNode1 = new ListNode(1);
        ListNode listNode2 = new ListNode(2);
        ListNode listNode3 = new ListNode(3);
        ListNode listNode4 = new ListNode(4);
        listNode1.next = listNode2;
        listNode2.next = listNode3;
        listNode3.next = listNode4;
        listNode4.next = new ListNode(5);
        reverseKGroup(listNode1,3);
    }
}