查找至少连续出现三次的所有数字/连续3天的日期【LeetCode】

编写一个SQL查询，查找至少连续出现三次的所有数字。
+----+-----+
| Id | Num |
+----+-----+
| 1  |  1  |
| 2  |  1  |
| 3  |  1  |
| 4  |  2  |
| 5  |  1  |
| 6  |  2  |
| 7  |  2  |
+----+-----+

【1】sql server优质解法：

【1.1】连续3次以上出现的数字

CREATE TABLE #A
(
    id INT IDENTITY(1,1),
    val INT
)
INSERT INTO #A(val) VALUES(1),(1),(1),(1),(2),(2),(3),(4),(4),(4),(4),(4)
INSERT INTO #A(val) VALUES(1)

select * from #A

SELECT val,MIN(id) AS minid,MAX(id) AS maxid, COUNT(1) AS cmd FROM
(
SELECT *,id-ROW_NUMBER() OVER( PARTITION BY val ORDER BY id )  AS idx FROM #A

) S GROUP BY val,idx

【1.2】连续出现的日期：（比如，想要查询连续登录超过3天的用户）

CREATE TABLE #b
(
id INT IDENTITY(1,1),
userid INT,
login_time datetime
)
INSERT INTO #b(userid,login_time)VALUES(101,''),(102,'')
INSERT INTO #b(userid,login_time)VALUES(102,'')
INSERT INTO #b(userid,login_time)VALUES(101,''),(102,'')
INSERT INTO #b(userid,login_time)VALUES(101,''),(102,'')
INSERT INTO #b(userid,login_time)VALUES(101,''),(102,'')
INSERT INTO #b(userid,login_time)VALUES(101,'')
INSERT INTO #b(userid,login_time)VALUES(101,'')
INSERT INTO #b(userid,login_time)VALUES(101,'')

SELECT * FROM #b ORDER BY userid ,login_time 

--解答
SELECT userid, MIN(login_time) AS StartDate, MAX(login_time) AS EndDate, COUNT(1) AS DayCount
FROM (
    SELECT userid
       ,login_time
       ,DATEADD(dd, -ROW_NUMBER() OVER ( PARTITION BY userid ORDER BY login_time), login_time) AS Grp
    FROM #b
) AS T
GROUP BY userid, [Grp]
ORDER BY 1

【2】mysql办法解决

【2.1】连续时间（比如，想要查询连续登录超过3天的用户）

（8.0以前，8.0以后可以用上述sql server 办法）

　测试代码

-- 测试数据代码
CREATE TABLE b
(
id INT primary key auto_increment,
userid INT,
login_time datetime
);
-- select * from b order by userid;
INSERT INTO b(userid,login_time) VALUES(101,''),(102,'');
INSERT INTO b(userid,login_time) VALUES(101,''),(102,'');
INSERT INTO b(userid,login_time) VALUES(101,''),(102,'');
INSERT INTO b(userid,login_time) VALUES(101,''),(102,'');
INSERT INTO b(userid,login_time) VALUES(101,''),(102,'');
INSERT INTO b(userid,login_time) VALUES(101,'');
INSERT INTO b(userid,login_time) VALUES(101,'');
INSERT INTO b(userid,login_time) VALUES(101,'');
INSERT INTO b(userid,login_time) VALUES(101,'');
INSERT INTO b(userid,login_time) VALUES(101,'');
INSERT INTO b(userid,login_time) VALUES(101,'');
INSERT INTO b(userid,login_time) VALUES(102,'');

插入后生成的测试表数据：　

　　　　

实现代码：

select userid,min(login_time) min_date,max(login_time) max_date,count(1) as day_count
from (
    select b.*,
    date_add(login_time,interval -if(@group_str=userid,@num:=@num+1,@num:=1) day) as login ,
    @group_str:=userid as temp
    from b cross join (select @num:=0,@group_str=-1) t
order by b.userid,login_time

) t
group by userid,login

结果：

【2.2】连续3次以上出现的数字

　　强烈推荐解法三

编写一个SQL查询，查找至少连续出现三次的所有数字。
+----+-----+
| Id | Num |
+----+-----+
| 1  |  1  |
| 2  |  1  |
| 3  |  1  |
| 4  |  2  |
| 5  |  1  |
| 6  |  2  |
| 7  |  2  |
+----+-----+

-- 建表
create table  Logs(id int primary key auto_increment,num int);

-- 造数据
INSERT INTO Logs(num) VALUES(1),(1),(1),(1),(2),(2),(3),(4),(4),(4),(4),(4),(1);

mysql解法：

这道题给了我们一个Logs表，让我们找Num列中连续出现相同数字三次的数字，那么由于需要找三次相同数字，所以我们需要建立三个表的实例.
我们可以用l1分别和l2, l3内交，l1和l2的Id下一个位置比，l1和l3的下两个位置比，然后将Num都相同的数字返回即可：

解法一:

SELECT DISTINCT l1.Num FROM Logs l1
JOIN Logs l2 ON l1.Id = l2.Id - 1
JOIN Logs l3 ON l1.Id = l3.Id - 2
WHERE l1.Num = l2.Num AND l2.Num = l3.Num;

　　

　　 

下面这种方法没用用到Join，而是直接在三个表的实例中查找，然后把四个条件限定上，就可以返回正确结果了：

解法二:

SELECT DISTINCT l1.Num 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;

　　再来看一种画风截然不同的方法，用到了变量count和pre，分别初始化为0和-1，然后需要注意的是用到了IF语句，MySQL里的IF语句和我们所熟知的其他语言的if不太一样，相当于我们所熟悉的三元操作符a?b:c，若a真返回b，否则返回c，具体可看这个帖子。那么我们先来看对于Num列的第一个数字1，pre由于初始化是-1，和当前Num不同，所以此时count赋1，此时给pre赋为1，然后Num列的第二个1进来，此时的pre和Num相同了，count自增1，到Num列的第三个1进来，count增加到了3，此时满足了where条件，t.n >= 3，所以1就被select出来了，以此类推遍历完整个Num就可以得到最终结果：

解法三：

SELECT DISTINCT Num FROM (
SELECT Num, @count := IF(@pre = Num, @count + 1, 1) AS n, @pre := Num
FROM Logs, (SELECT @count := 0, @pre := -1) AS init
) AS t WHERE t.n >= 3;