LeetCode——Rank Scores

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

此题，其本质就是赋值行号(需要注意分数相同的情景).

在实践过程中，初版答案如下所示：

# Write your MySQL query statement below
SELECT
    a.Score AS Score,
    (SELECT COUNT(DISTINCT b.Score) FROM Scores b where b.Score >= a.Score) AS Rank
FROM Scores a ORDER BY a.Score DESC;

此处，使用select count来统计行号，注意使用distinct来区分相同分数.
但是，此解题方案的效率较差，sql运行肯定是越快越好.
因此，在sql中引入变量来赋值行号，以替代耗时的select count操作.

# Write your MySQL query statement below
SELECT
    Score,
    @row := @row + (@pre <> (@pre := Score)) AS Rank
FROM Scores, (SELECT @row := 0, @pre := -1) t
ORDER BY Score DESC;

此处，查询是在Scores与临时表之间进行cross join.
此外，使用临时变量(@row,@pre)记录行号(此处，通过@pre与当前Score的比较，确定是否+1,需要注意mysql中的true/false为0/1).
通过以上改进，mysql的运行效率得到了较大的提高.