题目描述：

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

题目大意：

编写SQL对分数进行排序。如果两个分数相等，其排名应相同。注意在排名相等的分数之后，下一个排名的数值应该连续。换言之，排名之间不应该有“洞”（跳跃）。

例如给定的分数表，你的查询应该产生的结果如上所示（按照分数递减排序）。

解题思路：

使用MySQL的User-Defined Variables（用户定义变量）。

SQL语句：

# Write your MySQL query statement below
SELECT Score, Rank FROM(
  SELECT    Score,
            @curRank := @curRank + IF(@prevScore = Score, 0, 1) AS Rank,
            @prevScore := Score
  FROM      Scores s, (SELECT @curRank := 0) r, (SELECT @prevScore := NULL) p
  ORDER BY  Score DESC
) t;

另外，使用笛卡尔积也可以解此题。

参阅LeetCode Discuss：https://oj.leetcode.com/discuss/21473/accepted-solution-using-innerjoin-and-groupby

SELECT Scores.Score, COUNT(Ranking.Score) AS RANK
  FROM Scores
     , (
       SELECT DISTINCT Score
         FROM Scores
       ) Ranking
 WHERE Scores.Score <= Ranking.Score
 GROUP BY Scores.Id, Scores.Score
 ORDER BY Scores.Score DESC;

 

本文链接：http://bookshadow.com/weblog/2015/01/13/leetcode-rank-scores/
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