LeetCode（数据库）- 连续空余座位

题目链接：​​点击打开链接​​

题目大意：略。

解题思路：略。

AC 代码

-- 解题方案(1)
select distinct a.seat_id
from cinema a join cinema b
  on abs(a.seat_id - b.seat_id) = 1
  and a.free = true and b.free = true
order by a.seat_id;

-- 解题方案(2)
WITH t1 AS(SELECT c1.seat_id id1, c2.seat_id id2
FROM cinema c1
INNER JOIN cinema c2 ON c1.seat_id + 1 = c2.seat_id AND c1.free = 1 AND c2.free = 1)

SELECT *
FROM (SELECT id1 seat_id FROM t1
UNION 
SELECT id2 FROM t1) t
ORDER BY seat_id