LeetCode（数据库）- 游戏玩法分析 III

Posted 2021-05-29 Lux_Sun

题目链接：点击打开链接

 

题目大意：略。

 

解题思路：

• 解决方案(1)：自变量，算出每一行记录与上一行记录的累加即可
• 解决方案(2)：t1.event_date>=t2.event_date + GROUP BY t1.player_id,t1.event_date 是关键：（自连接）用 t1 比自己还小的所有在 t1.player_id,t1.event_date 范围内的时间，就是到达 t1 时间的总和游戏数

 

AC 代码

-- 解决方案(1)
SELECT RS.player_id, RS.event_date, RS.games_played_so_far
FROM (SELECT player_id, event_date, CONVERT(@sum:=IF(@p<>player_id, games_played, @sum + games_played), SIGNED) games_played_so_far, @p:=player_id
FROM Activity, (SELECT @sum:=0, @p:=-1) init
ORDER BY player_id, event_date) RS;

-- 解决方案(2)
SELECT t1.player_id,
       t1.event_date,
       SUM(t2.games_played) games_played_so_far
FROM Activity t1,Activity t2
WHERE t1.player_id=t2.player_id AND t1.event_date>=t2.event_date
GROUP BY t1.player_id,t1.event_date;