601.(Hard)体育馆的人流量

表：Stadium

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| visit_date    | date    |
| people        | int     |
+---------------+---------+
visit_date 是表的主键
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

编写一个 SQL 查询以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。

Stadium table:
+------+------------+-----------+
| 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       |
+------+------------+-----------+

Result table:
+------+------------+-----------+
| id   | visit_date | people    |
+------+------------+-----------+
| 5    | 2017-01-05 | 145       |
| 6    | 2017-01-06 | 1455      |
| 7    | 2017-01-07 | 199       |
| 8    | 2017-01-09 | 188       |
+------+------------+-----------+
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。

来源：力扣（LeetCode）

链接：https://leetcode-cn.com/problems/human-traffic-of-stadium

著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

Solution

1. 变量法

想复杂了，不过就当练习了。

with t1 as (
    SELECT *, (people>=100) busy
    FROM stadium
),t2 as (
    SELECT id, visit_date, people, busy,
        case 
        when @prev = busy then @slice := @slice
        when (@prev := busy) is not null then @slice := @slice+1 
        end flag
    from t1, (SELECT @prev:=null, @slice:=0) tmp
),t3 as (
select *, count(1) over (partition by flag) cnt from t2
)

SELECT id, visit_date, people from t3 
where busy =1 and cnt>=3 
order by 2

• t1 计算 busy

• t2 计算 flag（变量）

• t3 计算 cnt（窗口函数）

  id  visit_date   people  busy  flag cnt
  1   2017-01-01   10        0    1    1
  2   2017-01-02   109       1    2    2
  3   2017-01-03   150       1    2    2
  4   2017-01-04   99        0    3    1
  5   2017-01-05   145       1    4    4
  6   2017-01-06   1455      1    4    4
  7   2017-01-07   199       1    4    4
  8   2017-01-09   188       1    4    4

  事实上flag列的计算可以直接用 (id - row_number() OVER(ORDER BY id)) 结合 WHERE people >= 100 一步到位，如👇🏻窗口函数法

2.窗口函数法

with t1 as (
    SELECT  *
            ,(id - row_number() OVER(ORDER BY id)) flag
    FROM    stadium
    WHERE   people >= 100
), t2 as (
SELECT *,count(1) over (partition by flag) cnt from t1
)

SELECT id, visit_date, people from t2
where cnt>=3 order by 2

Table Schema

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');


-- case
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", '100');