力扣之1783.大满贯数量

news2025/1/12 4:43:45

文章目录

1. 1783.大满贯数量
- 1.1 题干
- 1.2 建表
- 1.3 题解
- 1.4 结果截图

1. 1783.大满贯数量

1.1 题干

表：Players

表：Championships

±--------------±--------+
| Column Name | Type |
±--------------±--------+
| year | int |
| Wimbledon | int |
| Fr_open | int |
| US_open | int |
| Au_open | int |
±--------------±--------+
year 是这个表的主键（具有唯一值的列）
该表的每一行都包含在每场大满贯网球比赛中赢得比赛的球员的 ID

编写解决方案，找出每一个球员赢得大满贯比赛的次数。结果不包含没有赢得比赛的球员的ID 。

结果集无顺序要求。

结果的格式，如下所示。

示例 1:

输入：
Players 表：
±----------±------------+
| player_id | player_name |
±----------±------------+
| 1 | Nadal |
| 2 | Federer |
| 3 | Novak |
±----------±------------+
Championships 表：
±-----±----------±--------±--------±--------+
| year | Wimbledon | Fr_open | US_open | Au_open |
±-----±----------±--------±--------±--------+
| 2018 | 1 | 1 | 1 | 1 |
| 2019 | 1 | 1 | 2 | 2 |
| 2020 | 2 | 1 | 2 | 2 |
±-----±----------±--------±--------±--------+
输出：
±----------±------------±------------------+
| player_id | player_name | grand_slams_count |
±----------±------------±------------------+
| 2 | Federer | 5 |
| 1 | Nadal | 7 |
±----------±------------±------------------+
解释：
Player 1 (Nadal) 获得了 7 次大满贯：其中温网 2 次(2018, 2019), 法国公开赛 3 次 (2018, 2019, 2020), 美国公开赛 1 次 (2018)以及澳网公开赛 1 次 (2018) 。
Player 2 (Federer) 获得了 5 次大满贯：其中温网 1 次 (2020), 美国公开赛 2 次 (2019, 2020) 以及澳网公开赛 2 次 (2019, 2020) 。
Player 3 (Novak) 没有赢得，因此不包含在结果集中。

1.2 建表

Create table If Not Exists Players (player_id int, player_name varchar(20))
Create table If Not Exists Championships (year int, Wimbledon int, Fr_open int, US_open int, Au_open int)
Truncate table Players
insert into Players (player_id, player_name) values ('1', 'Nadal')
insert into Players (player_id, player_name) values ('2', 'Federer')
insert into Players (player_id, player_name) values ('3', 'Novak')
Truncate table Championships
insert into Championships (year, Wimbledon, Fr_open, US_open, Au_open) values ('2018', '1', '1', '1', '1')
insert into Championships (year, Wimbledon, Fr_open, US_open, Au_open) values ('2019', '1', '1', '2', '2')
insert into Championships (year, Wimbledon, Fr_open, US_open, Au_open) values ('2020', '2', '1', '2', '2')

1.3 题解

select p.player_id,
       p.player_name,
       sum(player_id=wimbledon)+sum(player_id=fr_open)+sum(player_id=us_open)+sum(player_id=au_open) as grand_slams_count
from players p,championships c
group by p.player_id, p.player_name
having grand_slams_count>0
;