LeetCode force buckle brush title database (178): Score ranking

topic

Write a SQL query to achieve the score rankings. If the two scores are identical, the two scores rank (Rank) the same. Note that in the bisecting a ranking should be the next consecutive integer values. In other words, there should be no "gap" between the ranking.

For example, according to the table given Scores, your query should return (by score from highest to lowest):

analysis

select Score as Score
from Scores;

#将选择出来的命名为a
select a.Score as Score
from Scores as a;

#排序
select a.Score as Score
from Scores as a
order by a.Score DESC;

#比较难的是第二部分。假设现在给你一个分数X，如何算出它的排名Rank呢？ 我们可以先提取出大于等于X的所有分数集合H，将H去重后的元素个数就是X的排名。比如你考了99分，但最高的就只有99分，那么去重之后集合H里就只有99一个元素，个数为1，因此你的Rank为1。 先提取集合H：
select b.Score 
from Scores as b
where b.Score >=X;

#我们要的是集合H去重之后的元素个数，因此升级为：
select count(distinct b.Score) from Scores b where b.Score >= X as Rank;

-- 而从结果的角度来看，第二部分的Rank是对应一部分的分数来的，所以这里的X就是上面的a.Score，
select count(distinct b.Score) from Scores b where b.Score >= a.Score as Rank;

-- 把两部分结合在一起为：
select a.Score  as Score,
(select count(distinct b.Score) from Scores as b where b.Score>=a.score)as Rank
from Scores as a
order by a.Score desc;

answer

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