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
(selectseat_id,seat_id-row_number() over(order by seat_id) difffrom(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
(selectseat_id,seat_id-row_number() over(order by seat_id) difffrom(select seat_id from Cinema where free=1) t1
) t2
where diff in
(selectdifffrom (selectseat_id,seat_id-row_number() over(order by seat_id) difffrom(select seat_id from Cinema where free=1) t1) t2group by diff having count(seat_id) >=2
)
因此,最终代码为:
select
seat_id
from
(selectseat_id,seat_id-row_number() over(order by seat_id) difffrom(select seat_id from Cinema where free=1) t1
) t2
where diff in
(selectdifffrom (selectseat_id,seat_id-row_number() over(order by seat_id) difffrom(select seat_id from Cinema where free=1) t1) t2group by diff having count(seat_id) >=2
)