LeetCode-178:分数排名

题目描述：

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”。

+----+-------+
| Id | Score |
+----+-------+
| 1  | 3.50  |
| 2  | 3.65  |
| 3  | 4.00  |
| 4  | 3.85  |
| 5  | 4.00  |
| 6  | 3.65  |
+----+-------+

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    |
+-------+------+

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‘);

解题思路：

　　oracle可以使用排名函数进行排名，因为名次之间不应该有“间隔”，所以要使用dense_rank()函数，这个函数排的名次是连续的，rank()函数会跳过（并列排名），mysql依然是使用自定义变量来进行排名。

解题方案：

　　oracle

select round(a.score,2) as score, b.rank
  from Scores a
  join (select a.*, rownum as rank
          from (select distinct a.score from Scores a order by a.score desc) a) b
    on a.score = b.score
 order by b.rank

　　mysql

SELECT
    ROUND(a.score, 2) AS score,
    b.rank
FROM
    Scores a
JOIN (
    SELECT
        a.*, (@rowNum :=@rowNum + 1) AS rank
    FROM
        (
            SELECT DISTINCT
                a.score
            FROM
                Scores a
        ) a,
        (SELECT(@rowNum := 0)) b
    ORDER BY
        a.score DESC
) b ON a.score = b.score
ORDER BY
    b.rank

原文地址：https://www.cnblogs.com/zouqf/p/10283410.html