单挑力扣（LeetCode）SQL题：180. 连续出现的数字（难度：中等）

表：Logs
+-------------+---------+
| Column Name | Type    |
+-------------+---------+
| id          | int     |
| num         | varchar |
+-------------+---------+
id 是这个表的主键。


编写一个 SQL 查询，查找所有至少连续出现三次的数字。
返回的结果表中的数据可以按 任意顺序 排列。
查询结果格式如下面的例子所示：


示例 1:
输入：
Logs 表：
+----+-----+
| Id | Num |
+----+-----+
| 1  | 1   |
| 2  | 1   |
| 3  | 1   |
| 4  | 2   |
| 5  | 1   |
| 6  | 2   |
| 7  | 2   |
+----+-----+
输出：
Result 表：
+-----------------+
| ConsecutiveNums |
+-----------------+
| 1               |
+-----------------+
解释：1 是唯一连续出现至少三次的数字。


来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/consecutive-numbers

#测试数据
Create table If Not Exists Logs (id int, num int);


insert into Logs (id, num) values ('1', '1');
insert into Logs (id, num) values ('2', '1');
insert into Logs (id, num) values ('3', '1');
insert into Logs (id, num) values ('4', '2');
insert into Logs (id, num) values ('5', '1');
insert into Logs (id, num) values ('6', '2');

#方法1
select
    distinct b.num ConsecutiveNums
from (
    select  
        a.num,
        (row_number() over (order by a.id asc) - 
        row_number() over (partition by a.num order by a.id asc)) as series_id
    from logs a
) b
group by b.num, b.series_id
having count(1) >= 3;


#方法2
with tmp as (
select 
    a.id,a.num,
    row_number() over(order by a.id) rn
from Logs a
where (a.id,a.num) not in (select b.id+1,b.num from Logs b)
or (a.id,a.num) not in (select c.id-1,c.num from Logs c)
)
select
    a.num ConsecutiveNums
from tmp a
inner join tmp b
on a.rn = b.rn-1
and a.num = b.num
and b.id - a.id >=2
group by a.num;