601. 体育馆的人流量
原题: 601. 体育馆的人流量
困难
表：Stadium
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| id | int |
| visit_date | date |
| people | int |
+---------------+---------+
visit_date 是该表中具有唯一值的列。
每日人流量信息被记录在这三列信息中：序号 (id)、日期 (visit_date)、 人流量 (people)
每天只有一行记录，日期随着 id 的增加而增加
编写解决方案找出每行的人数大于或等于 100 且 id 连续的三行或更多行记录。
返回按 visit_date 升序排列 的结果表。
查询结果格式如下所示。
示例 1:
输入：
Stadium 表:
+------+------------+-----------+
| id | visit_date | people |
+------+------------+-----------+
| 1 | 2017-01-01 | 10 |
| 2 | 2017-01-02 | 109 |
| 3 | 2017-01-03 | 150 |
| 4 | 2017-01-04 | 99 |
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-09 | 188 |
+------+------------+-----------+
输出：
+------+------------+-----------+
| id | visit_date | people |
+------+------------+-----------+
| 5 | 2017-01-05 | 145 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-09 | 188 |
+------+------------+-----------+
解释：
id 为 5、6、7、8 的四行 id 连续，并且每行都有 >= 100 的人数记录。
请注意，即使第 7 行和第 8 行的 visit_date 不是连续的，输出也应当包含第 8 行，因为我们只需要考虑 id 连续的记录。
不输出 id 为 2 和 3 的行，因为至少需要三条 id 连续的记录。
题解
我的八嘎代码
直接暴力 列转行(满足人数 且 id - 1 == id(主) == id + 1), 行又转列(一行mysql呀~直接开除!)
SELECT
aa.aid as id,
aa.avisit_date AS visit_date,
aa.apeople AS people
FROM
(
select
b.id as aid,
b.visit_date as avisit_date,
b.people as apeople,
c.id as bid,
c.visit_date as bvisit_date,
c.people as bpeople,
d.id as cid,
d.visit_date as cvisit_date,
d.people as cpeople
from
(
select
a.*
from
Stadium as a
where
a.people >= 100
) as b
join
(
select
a.*
from
Stadium as a
where
a.people >= 100
) as c
on
b.id + 1 = c.id
join
(
select
a.*
from
Stadium as a
where
a.people >= 100
) as d
on
c.id + 1 = d.id
) as aa
UNION
SELECT
bb.bid as id,
bb.bvisit_date AS visit_date,
bb.bpeople AS people
FROM
(
select
b.id as aid,
b.visit_date as avisit_date,
b.people as apeople,
c.id as bid,
c.visit_date as bvisit_date,
c.people as bpeople,
d.id as cid,
d.visit_date as cvisit_date,
d.people as cpeople
from
(
select
a.*
from
Stadium as a
where
a.people >= 100
) as b
join
(
select
a.*
from
Stadium as a
where
a.people >= 100
) as c
on
b.id + 1 = c.id
join
(
select
a.*
from
Stadium as a
where
a.people >= 100
) as d
on
c.id + 1 = d.id
) as bb
UNION
SELECT
cc.cid as id,
cc.cvisit_date AS visit_date,
cc.cpeople AS people
FROM
(
select
b.id as aid,
b.visit_date as avisit_date,
b.people as apeople,
c.id as bid,
c.visit_date as bvisit_date,
c.people as bpeople,
d.id as cid,
d.visit_date as cvisit_date,
d.people as cpeople
from
(
select
a.*
from
Stadium as a
where
a.people >= 100
) as b
join
(
select
a.*
from
Stadium as a
where
a.people >= 100
) as c
on
b.id + 1 = c.id
join
(
select
a.*
from
Stadium as a
where
a.people >= 100
) as d
on
c.id + 1 = d.id
) as cc
ORDER BY id;
正解
在公司用hive常会用到这解法，有时候会在临时表里再多重嵌套
这道题需要提前用With临时空间，是因为where子句中需要再次调用from中选取的表
这里再聊一下sql的运行顺序：
- from -> where -> group by -> select -> order by -> limit
即临时表t1需要再from和where中都用到因此需要提前定义
天才!:
with t1 as(
select
*,
id - row_number() over(order by id) as rk -- 先满足people >= 100再标上序号
from stadium
where people >= 100
)
select id,visit_date,people
from t1
where rk in(
select rk
from t1
group by rk -- 按照序号分组, (为什么可以分组, 因为相差相同(见下面))
having count(rk) >= 3 -- 大于三个
)
-- 作者：宇航员
-- 链接：https://leetcode.cn/problems/human-traffic-of-stadium/solutions/701681/tu-jie-lian-xu-ri-qi-ji-nan-dian-fen-xi-xnj58/
-- 来源：力扣（LeetCode）
-- 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
示例
输入:
| id | visit_date | people |
| -- | ---------- | ------ |
| 1 | 2017-01-01 | 10 |
| 2 | 2017-01-02 | 109 |
| 3 | 2017-01-03 | 150 |
| 4 | 2017-01-04 | 999 |
| 5 | 2017-01-05 | 0 |
| 6 | 2017-01-06 | 1455 |
| 7 | 2017-01-07 | 199 |
| 8 | 2017-01-09 | 188 |
| 9 | 2024-03-14 | 100 |
代码:
select
*,
id as `|`,
row_number() over(order by id) as c ,
id - row_number() over(order by id) as rk
from stadium
where people >= 100
得:
| id | visit_date | people | | | c | rk |
| -- | ---------- | ------ | - | - | -- |
| 2 | 2017-01-02 | 109 | 2 | 1 | 1 |
| 3 | 2017-01-03 | 150 | 3 | 2 | 1 |
| 4 | 2017-01-04 | 999 | 4 | 3 | 1 |
| 6 | 2017-01-06 | 1455 | 6 | 4 | 2 |
| 7 | 2017-01-07 | 199 | 7 | 5 | 2 |
| 8 | 2017-01-09 | 188 | 8 | 6 | 2 |
| 9 | 2024-03-14 | 100 | 9 | 7 | 2 |
