LeetCode.601. 體育館的人流量

題目

SQL架構

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', '2017-01-08', '188')

X 市建了一個新的體育館，每日人流量信息被記錄在這三列信息中：序號 (id)、日期 (visit_date)、 人流量 (people)。

請編寫一個查詢語句，找出人流量的高峯期。高峯期時，至少連續三行記錄中的人流量不少於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       |
+------+------------+-----------+

對於上面的示例數據，輸出爲：

+------+------------+-----------+
| 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：
每天只有一行記錄，日期隨着 id 的增加而增加。

題解

如果只是找出全部人流量不少於100的記錄不難，難點在於如何查找連續的三天，一個想法是，查 3 張表，讓三個結果 id 連續

SELECT a.*
FROM stadium as a,stadium as b,stadium as c
where (a.id = b.id-1 and b.id+1 = c.id) 
  and (a.people>=100 and b.people>=100 and c.people>=100);

但是這樣輸出會有問題，比如 5,6,7,8 號人流量不少於100，但是隻輸出了 5,6號，根本原因在於，我們將 a 的 id 設爲三個連續值中最小值，所以只返回了每 3 個連續值中最小的一個，同理可想到，我們再將 a 的 id 設爲三個連續值中中間值和最大值，可以得到全部的連續 3 個值

SELECT a.*
FROM stadium as a,stadium as b,stadium as c
where ((a.id = b.id-1 and b.id+1 = c.id) or
       (a.id-1 = b.id and a.id+1 = c.id) or
       (a.id-1 = c.id and c.id-1 = b.id))
  and (a.people>=100 and b.people>=100 and c.people>=100);

但是這樣還有個問題，比如 5,6,7,8，6 既是 5,6,7 的中間值也是 6,7,8 的最小值，所以還要去重，也許 id 不按序排列，再排序 id，最終得到答案

SELECT distinct a.*
FROM stadium as a,stadium as b,stadium as c
where ((a.id = b.id-1 and b.id+1 = c.id) or
       (a.id-1 = b.id and a.id+1 = c.id) or
       (a.id-1 = c.id and c.id-1 = b.id))
  and (a.people>=100 and b.people>=100 and c.people>=100)
order by a.id;