原題:https://leetcode-cn.com/problems/rising-temperature/
我自己做leetcode的數據庫題目的解題記錄:
解題目錄 https://blog.csdn.net/weixin_42845682/article/details/105196004
題目
給定一個 Weather 表,編寫一個 SQL 查詢,來查找與之前(昨天的)日期相比溫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,根據上述給定的 Weather 表格,返回如下 Id:
+----+
| Id |
+----+
| 2 |
| 4 |
+----+
答案
第一種答案 - join
我以爲日期和id都是升序的,所以sql如下:
select
w2.id id
from weather w1
join weather w2 on w1.id = w2.id-1
where w2.temperature > w1.temperature
後來發現,有可能id升序,日期降序。所以改成如下:
select
w2.id id
from weather w1
join weather w2 on w1.recordDate = w2.recordDate-1
where w2.temperature > w1.temperature
但是報錯了…可我記得可以直接用加減操作日期啊…
換成函數,果然對了…
select
w2.id id
from weather w1
join weather w2 on DATEDIFF(w2.recordDate, w1.recordDate) = 1
where w2.temperature > w1.temperature
第二種答案 - over()
其實這道題over()也能做,但是leetcode裏的mysql好像不支持over()(mysql8.0好像支持over()了),所以就用oracle做。
select
id
from(
select
id id,
recordDate d,
temperature t,
lag(temperature) over(order by recordDate) as tt,
lag(recordDate) over(order by recordDate) as td,
lag(id) over(order by recordDate) as tid
from weather
) tmp
where t>tt
然後報錯了,我看了一下,輸入的數據只有兩條,id是升序,但是日期根本不是前一天後一天的關係…這也可以???
改一下sql…
select
id
from(
select
id id,
recordDate d,
temperature t,
lag(temperature) over(order by recordDate) as tt,
lag(recordDate) over(order by recordDate) as td,
lag(id) over(order by recordDate) as tid
from weather
) tmp
where t>tt
and d-td=1
嗯,發現tid那行沒啥用啊…刪了刪了
select
id
from(
select
id id,
recordDate d,
temperature t,
lag(temperature) over(order by recordDate) as tt,
lag(recordDate) over(order by recordDate) as td
from weather
) tmp
where t>tt
and d-td=1
但是發現一個奇怪的問題:
有紅框的是有多餘的tid的sql執行的時間,沒有紅框的,是沒有tid的sql執行的時間。
怎麼執行了多餘的語句,消耗時間還少了那麼多???
總感覺leetcode對oracle不是支持的很好…
順便吐槽一下,你想了好多細節的時候,leetcode告訴你你想多了;你不想那麼多的時候,你就又會感慨出一堆:臥槽,這也可以。