180. Consecutive Numbers

 

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.

 

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;

 

参考资料:

https://leetcode.com/discuss/54463/simple-solution

https://leetcode.com/discuss/87854/simple-sql-with-join-1484-ms

https://leetcode.com/discuss/69767/two-solutions-inner-join-and-two-variables

 

LeetCode All in One 题目讲解汇总(持续更新中…)


转载请注明来源于 Grandyang 的博客 (grandyang.com),欢迎对文章中的引用来源进行考证,欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论,也可以邮件至 grandyang@qq.com

💰


微信打赏


Venmo 打赏

×

Help us with donation