sql -leetcode 178. Rank Scores

Posted 2020-09-30 哈哈哈

Score 很好得到： select Score from Scores order by Score desc;

要得到rank， 可以通过比较比当前Score 大的Score 的个数得到：

select Score, (select count(distinct Score) from Scores where Score>=s.Score) Rank where Scores s order by Score desc;

　　

或者：

select s.Score ,count(distinct t.Score)Rank from Scores s join Scores t on s.Score<=t.Score

　　得到join 两个表，

{"headers": ["Score", "Score"], "values": [[3.50, 3.50], [3.50, 3.65], [3.65, 3.65], [3.65, 3.65], [3.50, 4.00], [3.65, 4.00], [4.00, 4.00], [3.85, 4.00], [4.00, 4.00], [3.65, 4.00], [3.50, 3.85], [3.65, 3.85], [3.85, 3.85], [3.65, 3.85], [3.50, 4.00], [3.65, 4.00], [4.00, 4.00], [3.85, 4.00], [4.00, 4.00], [3.65, 4.00], [3.50, 3.65], [3.65, 3.65], [3.65, 3.65]]}

按s.Id 聚在一起：

select s.Score ,count(distinct t.Score)Rank from Scores s join Scores t on s.Score<=t.Score
group by s.Id
order by s.Score desc