1841. 联赛信息统计（难度：中等）

题目描述

（通过次数2,430 | 提交次数4,581，通过率53.05%）

表: Teams
+----------------+---------+
| Column Name    | Type    |
+----------------+---------+
| team_id        | int     |
| team_name      | varchar |
+----------------+---------+
team_id 是该表主键.
每一行都包含了一个参加联赛的队伍信息.
表: Matches
+-----------------+---------+
| Column Name     | Type    |
+-----------------+---------+
| home_team_id    | int     |
| away_team_id    | int     |
| home_team_goals | int     |
| away_team_goals | int     |
+-----------------+---------+
(home_team_id, away_team_id) 是该表主键.
每一行包含了一次比赛信息.
home_team_goals 代表主场队得球数.
away_team_goals 代表客场队得球数.
获得球数较多的队伍为胜者队伍.

写一段SQL，用来报告联赛信息. 统计数据应使用已进行的比赛来构建，其中 获胜 球队获得 三分 ，而失败球队获得 零分。如果 打平 ，两支球队都得一分。
result 表的每行应包含以下信息:
team_name - Teams 表中的队伍名字
matches_played - 主场与客场球队进行的比赛次数.
points - 球队获得的总分数.
goal_for - 球队在所有比赛中获取的总进球数
goal_against - 球队在所有比赛中，他的对手球队的所有进球数
goal_diff - goal_for - goal_against.
按 points 降序 返回结果表。如果两队或多队得分相同，则按 goal_diff 降序 排列。如果仍然存在平局，则以team_name 按字典顺序 排列它们。

查询的结果格式如下例所示。
示例 1:
输入：
Teams 表:
+---------+-----------+
| team_id | team_name |
+---------+-----------+
| 1       | Ajax      |
| 4       | Dortmund  |
| 6       | Arsenal   |
+---------+-----------+
Matches 表:
+--------------+--------------+-----------------+-----------------+
| home_team_id | away_team_id | home_team_goals | away_team_goals |
+--------------+--------------+-----------------+-----------------+
| 1            | 4            | 0               | 1               |
| 1            | 6            | 3               | 3               |
| 4            | 1            | 5               | 2               |
| 6            | 1            | 0               | 0               |
+--------------+--------------+-----------------+-----------------+
输出：
+-----------+----------------+--------+----------+--------------+-----------+
| team_name | matches_played | points | goal_for | goal_against | goal_diff |
+-----------+----------------+--------+----------+--------------+-----------+
| Dortmund  | 2              | 6      | 6        | 2            | 4         |
| Arsenal   | 2              | 2      | 3        | 3            | 0         |
| Ajax      | 4              | 2      | 5        | 9            | -4        |
+-----------+----------------+--------+----------+--------------+-----------+
解释：
Ajax (team_id=1) 有4场比赛: 2败2平. 总分数 = 0 + 0 + 1 + 1 = 2.
Dortmund (team_id=4) 有2场比赛: 2胜. 总分数 = 3 + 3 = 6.
Arsenal (team_id=6) 有2场比赛: 2平. 总分数 = 1 + 1 = 2.
Dortmund 是积分榜上的第一支球队. Ajax和Arsenal 有同样的分数, 但Arsenal的goal_diff高于Ajax, 所以Arsenal在表中的顺序在Ajaxzhi'qian.

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/league-statistics

//测试数据
Create table If Not Exists Teams (team_id int, team_name varchar(20));
Create table If Not Exists Matches (home_team_id int, away_team_id int, home_team_goals int, away_team_goals int);

insert into Teams (team_id, team_name) values ('1', 'Ajax');
insert into Teams (team_id, team_name) values ('4', 'Dortmund');
insert into Teams (team_id, team_name) values ('6', 'Arsenal');

insert into Matches (home_team_id, away_team_id, home_team_goals, away_team_goals) values ('1', '4', '0', '1');
insert into Matches (home_team_id, away_team_id, home_team_goals, away_team_goals) values ('1', '6', '3', '3');
insert into Matches (home_team_id, away_team_id, home_team_goals, away_team_goals) values ('4', '1', '5', '2');
insert into Matches (home_team_id, away_team_id, home_team_goals, away_team_goals) values ('6', '1', '0', '0');

解题思路

Matches表中保存了每场比赛的主客场球队，以及对应的得分。
而题目要求，统计每支球队的参与、得分、进球情况。也就是说，不仅要统计球队作为主场的得分情况，也要统计球队作为客场的得分情况。
那么，我们可以将Matches表中的数据做转换，一行数据转换成2行数据，分别是主场球队视角以及对手球队的得分情况，以及客场球队视角以及对手球队的得分情况。
转换后，直接与Teams表关联，使用GROUP BY+SUM汇总统计即可。

参考SQL

未特别说明的情况下，参考SQL为基于MySQL8.0实现。

select
    a.team_name,
    count(1) matches_played,
    sum(case when b.team_goals > b.against_team_goals then 3
         when b.team_goals = b.against_team_goals then 1
         else 0 end) points,
    sum(b.team_goals) goal_for,
    sum(b.against_team_goals) goal_against,
    sum(b.team_goals-b.against_team_goals) goal_diff
from Teams a
inner join 
(
    select
        home_team_id team_id,home_team_goals team_goals,
        away_team_id against_team_id,away_team_goals against_team_goals 
    from Matches
    union all
    select
        away_team_id team_id,away_team_goals team_goals,
        home_team_id against_team_id,home_team_goals against_team_goals 
    from Matches
) b
on a.team_id = b.team_id
group by a.team_name
order by 3 desc,6 desc,1;