178. Rank Scores【leetcode】,sql

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

#--解法一 笛卡尔连接 分组计数 排序
SELECT Scores.Score, COUNT(Ranking.Score) AS RANK
  FROM Scores
     , (
       SELECT DISTINCT Score
         FROM Scores
       ) Ranking
#--等于是分组计数 每个分组中方的都是比自己大或相等的所有distinct元素
 WHERE Scores.Score <= Ranking.Score
# --分组
 GROUP BY Scores.Id, Scores.Score 
# --组id排序 降序
 ORDER BY Scores.Score DESC; 
 

#--解法二 mysql自定义变量
#-- Write your MySQL query statement below
SELECT Score, Rank 
FROM(
  SELECT    Score,
#--每一次都每一行都进行判断和计算
#--2 当值为Score,返回0, 否则返回1 也就是不重复的自增 重复的不变
            @curRank := @curRank + IF(@prevScore = Score, 0, 1) AS Rank,
            @prevScore := Score
#--1 初始变量值设置为0
  FROM      Scores s, (SELECT @curRank := 0) r, (SELECT @prevScore := NULL) p
#--3 降序排列
  ORDER BY  Score DESC
) t;