178. Rank Scores

最新推荐文章于 2024-12-31 10:28:59 发布

SnailRunFast

最新推荐文章于 2024-12-31 10:28:59 发布

阅读量362

点赞数

CC 4.0 BY-SA版权

分类专栏： Leet Code SQL 文章标签： Rank Scores

本文链接：https://blog.youkuaiyun.com/snailrunfast/article/details/70195708

Leet Code 同时被 2 个专栏收录

143 篇文章

订阅专栏

SQL

14 篇文章

订阅专栏

本文介绍了一个SQL查询案例，旨在解决分数排名的问题。当出现相同分数时，这些分数将获得相同的排名，并且排名不会出现断层。通过具体示例展示了如何使用SQL来实现这一功能。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Write a SQL query to rank scores. If there is a tie between two scores, both should have the same ranking. Note that after a tie, the next ranking number should be the next consecutive integer value. In other words, there should be no "holes" between ranks.

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

For example, given the above Scores table, your query should generate the following report (order by highest score):

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

Subscribe to see which companies asked this question.

Oracle : Partition over

MySQL:

# Write your MySQL query statement below


select
    s.Score  as Score,
    (select count(*) from (select Score from Scores group by Score) a where a.score >= s.score)   as Rank
from
    Scores s
order by
    s.Score desc
;