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.

Tips:

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);
在sql中,set/update语句中,=表示赋值；在set/update/select中,:=表示赋值，因此使用:=.

通过以上改进，mysql的运行效率得到了较大的提高.

PS:

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巴特西

LeetCode——Rank Scores

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