LeetCode数据库—分数排名

SQL架构：

Create table If Not Exists Scores (Id int, Score DECIMAL(3,2));
Truncate table Scores;
insert into Scores (Id, Score) values ('1', '3.5');
insert into Scores (Id, Score) values ('2', '3.65');
insert into Scores (Id, Score) values ('3', '4.0');
insert into Scores (Id, Score) values ('4', '3.85');
insert into Scores (Id, Score) values ('5', '4.0');
insert into Scores (Id, Score) values ('6', '3.65');

查看完整表记录

mysql> select * from scores;
+------+-------+
| Id   | Score |
+------+-------+
|    1 |  3.50 |
|    2 |  3.65 |
|    3 |  4.00 |
|    4 |  3.85 |
|    5 |  4.00 |
|    6 |  3.65 |
+------+-------+
6 rows in set (0.00 sec)

编写一个 SQL 查询来实现分数排名。如果两个分数相同，则两个分数排名（Rank）相同。请注意，平分后的下一个名次应该是下一个连续的整数值。换句话说，名次之间不应该有“间隔”。
例如，根据上述给定的 Scores 表，你的查询应该返回（按分数从高到低排列）：

+-------+------+
| Score | Rank |
+-------+------+
| 4.00  | 1    |
| 4.00  | 1    |
| 3.85  | 2    |
| 3.65  | 3    |
| 3.65  | 3    |
| 3.50  | 4    |
+-------+------+

解法

mysql> select score,
    -> (select count(distinct score) from scores where score>=s.score) Rank
    -> from scores s
    -> order by score desc;
+-------+------+
| score | Rank |
+-------+------+
|  4.00 |    1 |
|  4.00 |    1 |
|  3.85 |    2 |
|  3.65 |    3 |
|  3.65 |    3 |
|  3.50 |    4 |
+-------+------+
6 rows in set (0.00 sec)

这里可以延伸出另外一种排序方法
如下：
重复值的排名相同，但计次

+-------+------+
| score | rank |
+-------+------+
|  4.00 |    1 |
|  4.00 |    1 |
|  3.85 |    3 |
|  3.65 |    4 |
|  3.65 |    4 |
|  3.50 |    6 |
+-------+------+

解法

mysql> select score,
    -> (select count(score)+1 from scores s1 where s1.score>s2.score)as rank
    -> from scores s2
    -> order by score desc;
+-------+------+
| score | rank |
+-------+------+
|  4.00 |    1 |
|  4.00 |    1 |
|  3.85 |    3 |
|  3.65 |    4 |
|  3.65 |    4 |
|  3.50 |    6 |
+-------+------+
6 rows in set (0.00 sec)