文章目录
- 1. 题目
- 2. 解题
1. 题目
表: Scores
+---------------+---------+
| Column Name | Type |
+---------------+---------+
| player_name | varchar |
| gender | varchar |
| day | date |
| score_points | int |
+---------------+---------+
(gender, day)是该表的主键
一场比赛是在女队和男队之间举行的
该表的每一行表示一个名叫 (player_name) 性别为 (gender) 的参赛者在某一天获得了 (score_points) 的分数
如果参赛者是女性,那么 gender 列为 ‘F’,
如果参赛者是男性,那么 gender 列为 ‘M’
写一条SQL语句查询每种性别在每一天的总分,并按性别和日期对查询结果排序
下面是查询结果格式的例子:
Scores表:
+-------------+--------+------------+--------------+
| player_name | gender | day | score_points |
+-------------+--------+------------+--------------+
| Aron | F | 2020-01-01 | 17 |
| Alice | F | 2020-01-07 | 23 |
| Bajrang | M | 2020-01-07 | 7 |
| Khali | M | 2019-12-25 | 11 |
| Slaman | M | 2019-12-30 | 13 |
| Joe | M | 2019-12-31 | 3 |
| Jose | M | 2019-12-18 | 2 |
| Priya | F | 2019-12-31 | 23 |
| Priyanka | F | 2019-12-30 | 17 |
+-------------+--------+------------+--------------+
结果表:
+--------+------------+-------+
| gender | day | total |
+--------+------------+-------+
| F | 2019-12-30 | 17 |
| F | 2019-12-31 | 40 |
| F | 2020-01-01 | 57 |
| F | 2020-01-07 | 80 |
| M | 2019-12-18 | 2 |
| M | 2019-12-25 | 13 |
| M | 2019-12-30 | 26 |
| M | 2019-12-31 | 29 |
| M | 2020-01-07 | 36 |
+--------+------------+-------+
女性队伍:
第一天是 2019-12-30,Priyanka 获得 17 分,队伍的总分是 17 分
第二天是 2019-12-31, Priya 获得 23 分,队伍的总分是 40 分
第三天是 2020-01-01, Aron 获得 17 分,队伍的总分是 57 分
第四天是 2020-01-07, Alice 获得 23 分,队伍的总分是 80 分
男性队伍:
第一天是 2019-12-18, Jose 获得 2 分,队伍的总分是 2 分
第二天是 2019-12-25, Khali 获得 11 分,队伍的总分是 13 分
第三天是 2019-12-30, Slaman 获得 13 分,队伍的总分是 26 分
第四天是 2019-12-31, Joe 获得 3 分,队伍的总分是 29 分
第五天是 2020-01-07, Bajrang 获得 7 分,队伍的总分是 36 分
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/running-total-for-different-genders
著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。
2. 解题
类似题目:LeetCode MySQL 1204. 最后一个能进入电梯的人(累加/变量/窗口函数)
# Write your MySQL query statement below
select s1.gender, s1.day, sum(s2.score_points) total
from Scores s1, Scores s2
where s1.gender = s2.gender and s1.day >= s2.day
group by s1.day, s1.gender
order by s1.gender, s1.day
1104 ms
- 变量法
# Write your MySQL query statement below
select gender, day, case gender when 'F'then @prePointF := @prePointF + score_pointselse @prePointM := @prePointM + score_pointsend as total
from Scores, (select @prePointF := 0, @prePointM := 0) temp
order by gender, day
466 ms
- 窗口函数
# Write your MySQL query statement below
select gender, day, sum(score_points) over (partition by gender order by day) total
from Scores
order by gender, day
549 ms
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