LeetCode MySQL 601. 体育馆的人流量（row

LeetCode MySQL 601. 体育馆的人流量（row_number+over+cast）

文章目录

- 1. 题目
- 2. 解题

1. 题目

X 市建了一个新的体育馆，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)。

请编写一个查询语句，找出人流量的高峰期。高峰期时，至少连续三行记录中的人流量不少于100。

例如，表 stadium：

+------+------------+-----------+
| 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    | 2017-01-08 | 188       |
+------+------------+-----------+

对于上面的示例数据，输出为：

+------+------------+-----------+
| id   | visit_date | people    |
+------+------------+-----------+
| 5    | 2017-01-05 | 145       |
| 6    | 2017-01-06 | 1455      |
| 7    | 2017-01-07 | 199       |
| 8    | 2017-01-08 | 188       |
+------+------------+-----------+

提示：
每天只有一行记录，日期随着 id 的增加而增加。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/human-traffic-of-stadium
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2. 解题

使用 id 跟排序行号做差，连续的做差是一样的

select stadium.*, id - cast(row_number() over(partition by people >= 100) as signed) rnk 
from stadium
where people >= 100

{"headers": ["id", "visit_date", "people", "rnk"], 
"values": [
[2, "2017-01-02", 109, 1], 
[3, "2017-01-03", 150, 1], 
[5, "2017-01-05", 145, 2], 
[6, "2017-01-06", 1455, 2], 
[7, "2017-01-07", 199, 2], 
[8, "2017-01-08", 188, 2]]}

再套一层，算出 rnk 一样的有多少个

select id, visit_date, people,count(*) over(partition by rnk) cntfrom
(select stadium.*, id - cast(row_number() over(partition by people >= 100) as signed) rnkfrom stadiumwhere people >= 100
) t

{"headers": ["id", "visit_date", "people", "cnt"], 
"values": [
[2, "2017-01-02", 109, 2], 
[3, "2017-01-03", 150, 2], 
[5, "2017-01-05", 145, 4], 
[6, "2017-01-06", 1455, 4], 
[7, "2017-01-07", 199, 4], 
[8, "2017-01-08", 188, 4]]}

最后筛选 cnt >= 3 的

# Write your MySQL query statement below
select id, visit_date, people
from
(select id, visit_date, people,count(*) over(partition by rnk) cntfrom(select stadium.*, id - cast(row_number() over(partition by people >= 100) as signed) rnkfrom stadiumwhere people >= 100) t
) t
where cnt >= 3

或者 3表连接

# Write your MySQL query statement below
select distinct a.*
from stadium a, stadium b, stadium c
where ((b.id = a.id+1 and c.id = b.id+1) or(c.id = b.id+1 and a.id = c.id+1) or(a.id = c.id+1 and b.id = a.id+1))and a.people>=100 and b.people>=100 and c.people>=100
order by a.id