LeetCode（601）：体育馆的人流量 Human Traffic of Stadium（SQL）

2021.1.1 第一天，元旦快乐

LeetCode 从零单刷个人笔记整理（持续更新）

github：https://github.com/ChopinXBP/LeetCode-Babel

传送门：体育馆的人流量

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。

---

SELECT DISTINCT s1.id, s1.visit_date, s1.people
FROM stadium s1, stadium s2, stadium s3
WHERE s1.people >= 100 AND s2.people >= 100 AND s3.people >= 100 AND (
    ((s1.id = s2.id - 1) AND (s1.id = s3.id - 2))
    OR ((s1.id = s2.id - 1) AND (s1.id = s3.id + 1))
    OR ((s1.id = s2.id + 1) AND (s1.id = s3.id + 2))
)
ORDER BY s1.id;


SELECT id, visit_date, people
FROM (
	SELECT r1.*, @flag := if((r1.countt >= 3 OR @flag = 1) AND r1.countt != 0, 1, 0) AS flag
	FROM (
		SELECT s.*, @count := if(s.people >= 100, @count + 1, 0) AS `countt`
		FROM stadium s, (SELECT @count := 0) b
	) r1, (SELECT @flag := 0) c
	ORDER BY id DESC
) result
WHERE flag = 1 ORDER BY id;

---

#Coding一小时，Copying一秒钟。留个言点个赞呗，谢谢你#