SQL实现LeetCode(178.分数排行)

[LeetCode] 178.Rank Scores 分数排行

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    |
+-------+------+

这道题给了我们一个分数表,让我们给分数排序,要求是相同的分数在相同的名次,下一个分数在相连的下一个名次,中间不能有空缺数字,这道题我是完全照着史蒂芬大神的帖子来写的,膜拜大神中...大神总结了四种方法,那么我们一个一个的来膜拜学习,首先看第一种解法,解题的思路是对于每一个分数,找出表中有多少个大于或等于该分数的不同的分数,然后按降序排列即可,参见代码如下:

解法一:

SELECT Score,
(SELECT COUNT(DISTINCT Score) FROM Scores WHERE Score >= s.Score) Rank
FROM Scores s ORDER BY Score DESC;

跟上面的解法思想相同,就是写法上略有不同:

解法二:

SELECT Score,
(SELECT COUNT(*) FROM (SELECT DISTINCT Score s FROM Scores) t WHERE s >= Score) Rank
FROM Scores ORDER BY Score DESC;

下面这种解法使用了内交,Join是Inner Join的简写形式,自己和自己内交,条件是右表的分数大于等于左表,然后群组起来根据分数的降序排列,十分巧妙的解法:

解法三:

SELECT s.Score, COUNT(DISTINCT t.Score) Rank
FROM Scores s JOIN Scores t ON s.Score <= t.Score
GROUP BY s.Id ORDER BY s.Score DESC;

下面这种解法跟上面三种的画风就不太一样了,这里用了两个变量,变量使用时其前面需要加@,这里的:= 是赋值的意思,如果前面有Set关键字,则可以直接用=号来赋值,如果没有,则必须要使用:=来赋值,两个变量rank和pre,其中rank表示当前的排名,pre表示之前的分数,下面代码中的<>表示不等于,如果左右两边不相等,则返回true或1,若相等,则返回false或0。初始化rank为0,pre为-1,然后按降序排列分数,对于分数4来说,pre赋为4,和之前的pre值-1不同,所以rank要加1,那么分数4的rank就为1,下面一个分数还是4,那么pre赋值为4和之前的4相同,所以rank要加0,所以这个分数4的rank也是1,以此类推就可以计算出所有分数的rank了。

解法四:

SELECT Score,
@rank := @rank + (@pre <> (@pre := Score)) Rank
FROM Scores, (SELECT @rank := 0, @pre := -1) INIT
ORDER BY Score DESC;

参考资料:

https://leetcode.com/discuss/40116/simple-short-fast

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