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。 第一個玩家和第二個玩家的分數分別包含第一個玩家和第二個玩家的分數。 你可以假設，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得分最高的選手。
如果平局，player_id 最小 的選手獲勝。

編寫一個 SQL 查詢來查找每組中的獲勝者。

查詢結果格式如下所示

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

• 先把分數加起來

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
• 窗口函數獲取 組內排名

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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