Leetcode(Easy) - Rising Temperature

SQL Schema

Create table If Not Exists Weather (id int, recordDate date, temperature int)
Truncate table Weather
insert into Weather (id, recordDate, temperature) values ('1', '2015-01-01', '10')
insert into Weather (id, recordDate, temperature) values ('2', '2015-01-02', '25')
insert into Weather (id, recordDate, temperature) values ('3', '2015-01-03', '20')
insert into Weather (id, recordDate, temperature) values ('4', '2015-01-04', '30')

Table: Weather

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| id            | int     |
| recordDate    | date    |
| temperature   | int     |
+---------------+---------+
In SQL, id is the primary key for this table.
This table contains information about the temperature on a certain day.

Question

Find all dates’ Id with higher temperatures compared to its previous dates (yesterday).

Return the result table in any order.

The result format is in the following example.

<Example 1>

Input:
Weather table:
+----+------------+-------------+
| id | recordDate | temperature |
+----+------------+-------------+
| 1  | 2015-01-01 | 10          |
| 2  | 2015-01-02 | 25          |
| 3  | 2015-01-03 | 20          |
| 4  | 2015-01-04 | 30          |
+----+------------+-------------+
Output:
+----+
| id |
+----+
| 2  |
| 4  |
+----+
Explanation:
In 2015-01-02, the temperature was higher than the previous day (10 -> 25).
In 2015-01-04, the temperature was higher than the previous day (20 -> 30).

Solution

select td.id
from Weather td
join Weather yd
on td.recordDate = DATE_ADD(yd.recordDate, interval 1 day)
where td.temperature > yd.temperature