Write a SQL query to find all numbers that appear at least three times consecutively.
+----+-----+
| Id | Num |
+----+-----+
| 1 | 1 |
| 2 | 1 |
| 3 | 1 |
| 4 | 2 |
| 5 | 1 |
| 6 | 2 |
| 7 | 2 |
+----+-----+
For example, given the above Logs table, 1 is the only number that appears consecutively for at least three times.
使用自连接，然后一一和上一个比较，最后分组过滤即可
select log1.Num from Logs log1
join Logs log2
on log1.Id + 1 = log2.Id and log1.Num = log2.Num
group by log1.Num
having count(log1.Num) >= 2;结果超时, 可能是分组导致的。
还是使用自连接，不过为了避免group开销，直接使用三个表连接，分别指向Id, Id + 1, Id + 2，并分别让Num相等即可
select distinct a.Num
from Logs a, Logs b,Logs c
where a.Id=b.Id+1 and a.Num=b.Num and b.Id=c.Id+1 and b.Num=c.Num;统计法。如果这一题使用c编程实现，逐行读取，这就容易了。
可以使用case语句实现:
select Num from (
SELECT
Num,
CASE
WHEN @prevNum = Num THEN @count := @count + 1
WHEN @prevNum := Num THEN @count := 1
END n
FROM Logs, (select @prevNum := NULL) initPrevNum, (select @count := 1) initCount
) as result where result.n >= 3;注意在case后的结果别名为n，当不能直接在该语句中使用，必须使用另一个select语句查询。
when是一个短路的，相当于if。:=用于赋值，=用于比较,必须在from语句中初始化prevNum, count的值
相当于switch语句，when后面结条件，并且是短路的。例子:
select Num,
case
when Num % 2 = 0 then "Even"
when Num % 2 = 1 then "Odd"
end Numtype
from Logs;