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 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;
参考资料:
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