sql求解连续两个以上的空座位

Q：查找电影院所有连续可用的座位。
返回按 seat_id 升序排序 的结果表。
测试用例的生成使得两个以上的座位连续可用。
结果表格式如下所示。

A:我们首先找出所有的空座位：1，3，4，5
按照seat_id排序（上面已经有序），并赋予排名，
则

seat_id	rnk	seta_id-rnk
1	1	0
3	2	1
4	3	1
5	4	1

我们发现连续的数与其对应的排名均是连续的，那么两者相减应该等于相同的数，因此可以分为以下几步

# 1. t1:获取所有空座位
select seat_id from Cinema where free=1

# 2. t2:获取所有的连续数字对应的组
select
seat_id,
seat_id-row_number() over(order by seat_id) diff
from
(
	select seat_id from Cinema where free=1
) t1

# 3. t3:连续的数对应的diff是相同的，可以按照diff分组，并收集空座位号大于等于2的组号
select
diff
from 
(
	select
	seat_id,
	seat_id-row_number() over(order by seat_id) diff
	from
		(
			select seat_id from Cinema where free=1
		) t1
) t2
group by diff having count(seat_id) >=2

# 4. 根据t2表以及t3表获取大于等于2个以上的空座位号
select 
seat_id
from 
(
	select
	seat_id,
	seat_id-row_number() over(order by seat_id) diff
	from
	(
		select seat_id from Cinema where free=1
	) t1
) t2
where diff in 
(
	select
	diff
	from 
	(
		select
		seat_id,
		seat_id-row_number() over(order by seat_id) diff
		from
			(
				select seat_id from Cinema where free=1
			) t1
	) t2
	group by diff having count(seat_id) >=2
) 

因此，最终代码为：

select 
seat_id
from 
(
	select
	seat_id,
	seat_id-row_number() over(order by seat_id) diff
	from
	(
		select seat_id from Cinema where free=1
	) t1
) t2
where diff in 
(
	select
	diff
	from 
	(
		select
		seat_id,
		seat_id-row_number() over(order by seat_id) diff
		from
			(
				select seat_id from Cinema where free=1
			) t1
	) t2
	group by diff having count(seat_id) >=2
)