MYSQL连续问题
问题一：连续空余座位
几个朋友来到电影院的售票处，准备预约连续空余座位。利用表 cinema ，写一个查询语句，获取所有空余座位，并将它们按照 seat_id 排序后返回。
注意：
- seat_id 字段是一个自增的整数，free 字段是布尔类型（‘1’ 表示空余， ‘0’ 表示已被占据）。
- 连续空余座位的定义是大于等于 2 个连续空余的座位。
| seat_id | free |
|---------|------|
| 1 | 1 |
| 2 | 0 |
| 3 | 1 |
| 4 | 1 |
| 5 | 1 |
解法一：自连接（使用Join 和 abs）
解题思路：
1）将表格自连接，得到两个表的笛卡尔乘积，如下图（注意： 这里只放了一部分）
放个百度百科的图片帮助理解：
2）为了找到连续空座位，a.seat_id 和b.seat_id应该是相邻的两个数用错位相减法on abs(a.seat_id - b.seat_id)=1； 且两者都应该为空and a.free = true and b.free = true
3）防止出现重复值要加distinct，比如34和45都满足条件，4会出现2次
select distinct a.seat_id
from cinema a join cinema b
on abs(a.seat_id - b.seat_id) = 1
and a.free = true and b.free = true
order by a.seat_id
解法二：自连接+where
解题思路：和解法一思路基本一致，改成用where 筛选。
select distinct c1.seat_id as seat_id
from cinema as c1,cinema as c2
where (c1.seat_id=c2.seat_id-1 or c1.seat_id=c2.seat_id+1)
and c1.free =1
and c2.free=1
order by seat_id
解法三：使用窗口函数
解题思路：使用Lag()和lead()函数
#第一种窗口函数
select seat_id
from (
select
seat_id ,
free ,
lag(free,1,999) over() pre_free,
lead(free,1,999) over() next_free
from cinema
)tmp
where free=1 and (pre_free=1 or next_free=1 )
order by seat_id
#第二种窗口函数
SELECT C.seat_id
FROM
(SELECT seat_id,
sum(free = 1) OVER (ORDER BY seat_id ASC rows 1 preceding) AS r1,
sum(free = 1) OVER (ORDER BY seat_id ASC rows between 1 preceding and 1 following) AS r2,
sum(free = 1) OVER (ORDER BY seat_id ASC rows between 0 preceding and 1 following) AS r3
FROM cinema) AS C
WHERE C.r1 = 2 OR C.r2 = 3 OR C.r3 = 2;
问题二：连续 5 天登录的用户
写一个 SQL 查询, 找到活跃用户的 id 和 name。活跃用户是指那些至少连续 5 天登录账户的用户。返回的结果表按照 id 排序.
后续问题:如果活跃用户是那些至少连续 n 天登录账户的用户, 你能否写出通用的解决方案?
Accounts 表:
+----+----------+
| id | name |
+----+----------+
| 1 | Winston |
| 7 | Jonathan |
+----+----------+
Logins 表:
+----+------------+
| id | login_date |
+----+------------+
| 7 | 2020-05-30 |
| 1 | 2020-05-30 |
| 7 | 2020-05-31 |
| 7 | 2020-06-01 |
| 7 | 2020-06-02 |
| 7 | 2020-06-02 |
| 7 | 2020-06-03 |
| 1 | 2020-06-07 |
| 7 | 2020-06-10 |
+----+------------+
解法1: 定义变量
SELECT
DISTINCT b.id,b.name
FROM
(SELECT
id,
CASE
WHEN @id = id
AND @prev_date=subdate(login_date, interval 1 day)
AND (@prev_date:= login_date) IS NOT NULL
THEN @cnt := @cnt+1
WHEN @id = id
AND @prev_date=subdate(login_date, interval 0 day)
THEN @cnt := @cnt
WHEN (@id := id)IS NOT NULL
AND(@prev_date:= login_date) IS NOT NULL
THEN @cnt:=1
END AS cnt
FROM
(SELECT * FROM logins ORDER BY id,login_date) a,
(SELECT @prev_date:=NULL, @cnt:= NULL, @id:= NULL) t) a
LEFT JOIN accounts b ON a.id = b.id
WHERE a.cnt>=5
解法2：窗口函数
解题思路：
1、原表GROUP BY 分组；
2、使用窗口函数ROW_NUMBER()添加序号；
3、计算登录日期与序号的差值，并以差值重新分组，筛选连续5天登录的用户；
4、对id去重；
5、LEFT JOIN 关联姓名即可。
SELECT t4.id, name
FROM
(
SELECT DISTINCT id
FROM
(
SELECT id, SUBDATE(login_date, rn) AS diff
FROM
( #对日期进行排序
SELEC id ,login_date ,ROW_NUMBER() OVER(PARTITION BY id ORDER BY login_date) AS rn
FROM
#获取去重的表格数据
(SELECT id, login_date FROM Logins GROUP BY id, login_date)t1
) t2
GROUP BY id, diff
HAVING COUNT(*) >= 5
) t3
) t4
JOIN Accounts t5 ON t4.id = t5.id
ORDER BY id
解法3：自连接
解题思路：
– DATEDIFF(L2.login_date,L1.login_date) BETWEEN 0 AND 4 ：对每个id相同的日期都只连接相差4天之内的日期。即对每个日期都检索一下是否有5个连续的日期。
– HAVING COUNT(DISTINCT L2.login_date)=5：只要5天之内，日期有一个不连续，那么去重后每一组的天数必然少于5天。
SELECT DISTINCT A.*
FROM Accounts A
JOIN logins L1 USING(id)
JOIN Logins L2 ON L1.id=L2.id AND DATEDIFF(L2.login_date,L1.login_date) BETWEEN 0 AND 4
GROUP BY A.id,A.name,L1.login_date
HAVING COUNT(DISTINCT L2.login_date)=5
题目三：连续出现的数字
编写一个 SQL 查询，查找所有至少连续出现三次的数字。
例如，给定的 Logs 表， 1 是唯一连续出现至少三次的数字。
+----+-----+
| Id | Num |
+----+-----+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 1 |
| 6 | 2 |
| 7 | 2 |
+----+-----+
解法一：直接方法
连续出现的意味着相同数字的 Id 是连着的，由于这题问的是至少连续出现 3 次，我们使用 Logs 并检查是否有 3 个连续的相同数字。
SELECT DISTINCT
l1.Num AS ConsecutiveNums
FROM
Logs l1,
Logs l2,
Logs l3
WHERE
l1.Id = l2.Id - 1
AND l2.Id = l3.Id - 1
AND l1.Num = l2.Num
AND l2.Num = l3.Num;
解法二：窗口函数
#1）使用LAG
SELECT DISTINCT Num as ConsecutiveNums
FROM(
SELECT Id,Num,
LAG(Num,1)OVER(ORDER BY Id) as num_1,
LAG(Num,2)OVER(ORDER BY Id) as num_2
FROM Logs
) as c
WHERE c.num = c.num1 and c.num1 = c.num2
# 2）也可以用LEAD替换LAG
select distinct num as ConsecutiveNums
from(
select num,
lead(num,1)over()as num1,
lead(num,2)over()as num2
from logs
) as c
where c.num = c.num1 and c.num1 = c.num2
解法三：变量
select distinct Num as ConsecutiveNums
from (
select Num,
case
when @prev = Num then @count := @count + 1
when (@prev := Num) is not null then @count := 1
end as CNT
from Logs, (select @prev := null,@count := null) as t
) as temp
where temp.CNT >= 3
题目一链接：https://leetcode-cn.com/problems/consecutive-available-seats
题目二链接：https://leetcode-cn.com/problems/active-users
题目三连接：https://leetcode-cn.com/problems/consecutive-numbers