leetcode 178

分数排名

select s1.score, count(distinct s2.score) as `rank`
from Scores as s1, Scores as s2
where s1.score <= s2.score
group by s1.id
order by s1.score desc

mysql 8.0 下新增窗口函数 dense_rank()

select score, dense_rank() over(order by score desc)  `rank`
from Scores ;