LeetCode_sql_day16（601.体育馆的人流量）

描述：601. 体育馆的人流量 - 力扣（LeetCode）

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。

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。

数据准备：

Create table If Not Exists Stadium (id int, visit_date DATE NULL, people int)

Truncate table Stadium

insert into Stadium (id, visit_date, people) values ('1', '2017-01-01', '10')

insert into Stadium (id, visit_date, people) values ('2', '2017-01-02', '109')

insert into Stadium (id, visit_date, people) values ('3', '2017-01-03', '150')

insert into Stadium (id, visit_date, people) values ('4', '2017-01-04', '99')

insert into Stadium (id, visit_date, people) values ('5', '2017-01-05', '145')

insert into Stadium (id, visit_date, people) values ('6', '2017-01-06', '1455')

insert into Stadium (id, visit_date, people) values ('7', '2017-01-07', '199')

insert into Stadium (id, visit_date, people) values ('8', '2017-01-09', '188')

分析：

整体思路和之前连续问题一样，构造等差数列解决连续问题

①先对整体排名

select *,row_number() over (order by visit_date)r1 from stadium

②对people大于等于100的再排个序

with t1 as (
select *,row_number() over (order by visit_date)r1 from stadium)
select *,row_number() over (order by visit_date)r2 from t1
where people >= 100

③对r1,r2两列做差,差值相同的即连续数列，同时用count求出连续天数

with t1 as (
select *,row_number() over (order by visit_date)r1 from stadium)
, t2 as(
select *,row_number() over (order by visit_date)r2 from t1
where people >= 100)
select *,r1-r2 as r3,count(r1-r2) over(partition by (r1-r2))r4 from t2

④筛选数据，并按照id排序

select id,visit_date,people
from t3
where r4 >=3
order by id

代码：

with t1 as (
select *,row_number() over (order by visit_date)r1 from stadium)
, t2 as(
select *,row_number() over (order by visit_date)r2 from t1
where people >= 100)
, t3 as (
select *,r1-r2 as r3,count(r1-r2) over(partition by (r1-r2))r4 from t2)
select id,visit_date,people
from t3
where r4 >=3
order by id;

总结：

连续问题记得使用两个排名函数构造等差数列