leetcode 197. 上升的温度

题目

给定一个 Weather 表，编写一个 SQL 查询，来查找与之前（昨天的）日期相比温度更高的所有日期的 Id。

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

例如，根据上述给定的 Weather 表格，返回如下 Id:

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

解法

这里我们用到SQL Server 里面的DATEDIFF函数，此函数返回在指定的startdate和enddate之间相差的指定的日期部分边界的计数。就是说你拿天数作比较，它会返回你的天数之差。

官方文档

DATEDIFF ( datepart , startdate , enddate )

例子

SELECT DATEDIFF(year, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000'); SELECT DATEDIFF(quarter, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000'); SELECT DATEDIFF(month, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000'); SELECT DATEDIFF(dayofyear, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000'); SELECT DATEDIFF(day, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000'); SELECT DATEDIFF(week, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000'); SELECT DATEDIFF(hour, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000'); SELECT DATEDIFF(minute, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000'); SELECT DATEDIFF(second, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000'); SELECT DATEDIFF(millisecond, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000'); SELECT DATEDIFF(microsecond, '2005-12-31 23:59:59.9999999', '2006-01-01 00:00:00.0000000');

代码

/* Write your T-SQL query statement below */ SELECT w2.id AS 'Id' FROM Weather w1, Weather w2 WHERE DATEDIFF(day, w1.RecordDate, w2.RecordDate) = 1 AND w1.Temperature < w2.Temperature;

结果