[LeetCode] Rising Temperature 上升温度

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[LeetCode] Rising Temperature 上升温度

李博 bluemind 2017-12-14 11:56:00 浏览741
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Given a Weather table, write a SQL query to find all dates' Ids with higher temperature compared to its previous (yesterday's) dates.

+---------+------------+------------------+
| Id(INT) | Date(DATE) | Temperature(INT) |
+---------+------------+------------------+
|       1 | 2015-01-01 |               10 |
|       2 | 2015-01-02 |               25 |
|       3 | 2015-01-03 |               20 |
|       4 | 2015-01-04 |               30 |
+---------+------------+------------------+

For example, return the following Ids for the above Weather table:

+----+
| Id |
+----+
|  2 |
|  4 |
+----+

这道题给了我们一个Weather表,让我们找出比前一天温度高的Id,由于Id的排列未必是按顺序的,所以我们要找前一天就得根据日期来找,我们可以使用MySQL的函数Datadiff来计算两个日期的差值,我们的限制条件是温度高且日期差1,参见代码如下: 

解法一:

SELECT w1.Id FROM Weather w1, Weather w2
WHERE w1.Temperature > w2.Temperature AND DATEDIFF(w1.Date, w2.Date) = 1;

下面这种解法我们使用了MySQL的TO_DAYS函数,用来将日期换算成天数,其余跟上面相同:

解法二:

SELECT w1.Id FROM Weather w1, Weather w2
WHERE w1.Temperature > w2.Temperature AND TO_DAYS(w1.Date) = TO_DAYS(w2.Date) + 1;

我们也可以使用Subdate函数,来实现日期减1,参见代码如下:

解法三:

SELECT w1.Id FROM Weather w1, Weather w2
WHERE w1.Temperature > w2.Temperature AND SUBDATE(w1.Date, 1) = w2.Date;

最后来一种完全不一样的解法,使用了两个变量pre_t和pre_d分别表示上一个温度和上一个日期,然后当前温度要大于上一温度,且日期差为1,满足上述两条件的话选出来为Id,否则为NULL,然后更新pre_t和pre_d为当前的值,最后选出的Id不为空即可:

解法四:

SELECT Id FROM (
SELECT CASE WHEN Temperature > @pre_t AND DATEDIFF(Date, @pre_d) = 1 THEN Id ELSE NULL END AS Id,
@pre_t := Temperature, @pre_d := Date 
FROM Weather, (SELECT @pre_t := NULL, @pre_d := NULL) AS init ORDER BY Date ASC
) id WHERE Id IS NOT NULL;

本文转自博客园Grandyang的博客,原文链接:上升温度[LeetCode] Rising Temperature ,如需转载请自行联系原博主。

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