SQL实现LeetCode(197.上升温度)

[LeetCode] 197.Rising Temperature 上升温度

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 w2WHERE w1.Temperature > w2.Temperature AND DATEDIFF(w1.Date, w2.Date) = 1;

解法二：

SELECT w1.Id FROM Weather w1, Weather w2WHERE w1.Temperature > w2.Temperature AND TO_DAYS(w1.Date) = TO_DAYS(w2.Date) + 1;

解法三：

SELECT w1.Id FROM Weather w1, Weather w2WHERE w1.Temperature > w2.Temperature AND SUBDATE(w1.Date, 1) = w2.Date;

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;

https://leetcode.com/discuss/33641/two-solutions

https://leetcode.com/discuss/52370/my-simple-solution-using-inner-join

https://leetcode.com/discuss/86435/a-simple-straightforward-solution-and-its-very-fast

[LeetCode] 197.Rising Temperature 上升温度

更多相关文章

随机推荐