LeetCode：Database 66.锦标赛优胜者

Posted 2021-06-26 Xiao Miao

要求：编写一个 SQL 查询来查找每组中的获胜者。

每组的获胜者是在组内累积得分最高的选手。如果平局，player_id 最小 的选手获胜。

Players 玩家表的结构：

+-------------+-------+
| Column Name | Type  |
+-------------+-------+
| player_id   | int   |
| group_id    | int   |
+-------------+-------+
player_id 是此表的主键。
此表的每一行表示每个玩家的组。

Matches 赛事表的结构：

+---------------+---------+
| Column Name   | Type    |
+---------------+---------+
| match_id      | int     |
| first_player  | int     |
| second_player | int     | 
| first_score   | int     |
| second_score  | int     |
+---------------+---------+
match_id 是此表的主键。
每一行是一场比赛的记录，first_player 和 second_player 表示该场比赛的球员 ID。
first_score 和 second_score 分别表示 first_player 和 second_player 的得分。
你可以假设，在每一场比赛中，球员都属于同一组。

Players 表:

+-----------+------------+
| player_id | group_id   |
+-----------+------------+
| 15        | 1          |
| 25        | 1          |
| 30        | 1          |
| 45        | 1          |
| 10        | 2          |
| 35        | 2          |
| 50        | 2          |
| 20        | 3          |
| 40        | 3          |
+-----------+------------+

Matches 表:

+------------+--------------+---------------+-------------+--------------+
| match_id   | first_player | second_player | first_score | second_score |
+------------+--------------+---------------+-------------+--------------+
| 1          | 15           | 45            | 3           | 0            |
| 2          | 30           | 25            | 1           | 2            |
| 3          | 30           | 15            | 2           | 0            |
| 4          | 40           | 20            | 5           | 2            |
| 5          | 35           | 50            | 1           | 1            |
+------------+--------------+---------------+-------------+--------------+

Result Table:

+-----------+------------+
| group_id  | player_id  |
+-----------+------------+ 
| 1         | 15         |
| 2         | 35         |
| 3         | 40         |
+-----------+------------+

SQL语法：

WITH b AS (SELECT first_player,SUM(first_score) AS s1 FROM matches GROUP BY first_player)
, c AS (SELECT second_player,SUM(second_score) AS s2 FROM matches GROUP BY second_player)


select group_id,player_id from(
select group_id,player_id,num,row_number() over(partition by group_id order by num desc,player_id asc) as r from(
select a.player_id,a.group_id,(ifnull(b.s1,0)+ifnull(c.s2,0)) as num
from players  a
left join  b
on a.player_id=b.first_player
left join  c
on c.second_player=a.player_id)d)e
where r=1;