文章目錄

• • 1. 題目
  • 2. 解題

1. 題目

Players 玩家表

+-------------+-------+ | Column Name | Type | +-------------+-------+ | player_id | int | | group_id | int | +-------------+-------+ 玩家 ID 是此表的主鍵。 此表的每一行表示每個(gè)玩家的組。

Matches 賽事表

+---------------+---------+ | Column Name | Type | +---------------+---------+ | match_id | int | | first_player | int | | second_player | int | | first_score | int | | second_score | int | +---------------+---------+ match_id 是此表的主鍵。 每一行是一場比賽的記錄，第一名和第二名球員包含每場比賽的球員 ID。 第一個(gè)玩家和第二個(gè)玩家的分?jǐn)?shù)分別包含第一個(gè)玩家和第二個(gè)玩家的分?jǐn)?shù)。 你可以假設(shè)，在每一場比賽中，球員都屬于同一組。

每組的獲勝者是在組內(nèi)得分最高的選手。
如果平局，player_id 最小 的選手獲勝。

編寫一個(gè) SQL 查詢來查找每組中的獲勝者。

查詢結(jié)果格式如下所示

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

來源：力扣（LeetCode） 鏈接：https://leetcode-cn.com/problems/tournament-winners
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2. 解題

• 先把分?jǐn)?shù)加起來

select player_id, sum(score) total_score from ((select first_player player_id, first_score scorefrom Matches)union all(select second_player, second_scorefrom Matches) ) t1 group by player_id {"headers": ["player_id", "total_score"], "values": [[15, 3], [30, 3], [40, 5], [35, 1], [45, 0], [25, 2], [20, 2], [50, 1]]}

• 左連接獲取 分組id
• 窗口函數(shù)獲取 組內(nèi)排名

select group_id, player_id,rank() over(partition by group_id order by total_score desc, player_id) rnk from (select group_id, t2.player_id, total_scorefrom Players p left join(select player_id, sum(score) total_scorefrom((select first_player player_id, first_score scorefrom Matches)union all(select second_player, second_scorefrom Matches)) t1group by player_id) t2using(player_id) ) t3

• 取出排名為1的

# Write your MySQL query statement belowselect group_id, player_id from (select group_id, player_id,rank() over(partition by group_id order by total_score desc, player_id) rnkfrom(select group_id, t2.player_id, total_scorefrom Players p left join(select player_id, sum(score) total_scorefrom((select first_player player_id, first_score scorefrom Matches)union all(select second_player, second_scorefrom Matches)) t1group by player_id) t2using(player_id)) t3 ) t where rnk = 1

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