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CyC2018
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@ -585,7 +585,10 @@ SELECT
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FROM
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Customers
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WHERE
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Id NOT IN ( SELECT CustomerId FROM Orders );
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Id NOT IN (
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SELECT CustomerId
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FROM Orders
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);
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```
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## SQL Schema
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@ -666,7 +669,9 @@ SELECT
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FROM
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Employee E,
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Department D,
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( SELECT DepartmentId, MAX( Salary ) Salary FROM Employee GROUP BY DepartmentId ) M
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( SELECT DepartmentId, MAX( Salary ) Salary
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FROM Employee
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GROUP BY DepartmentId ) M
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WHERE
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E.DepartmentId = D.Id
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AND E.DepartmentId = M.DepartmentId
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@ -727,7 +732,10 @@ https://leetcode.com/problems/second-highest-salary/description/
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```sql
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SELECT
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( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
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( SELECT DISTINCT Salary
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FROM Employee
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ORDER BY Salary DESC
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LIMIT 1, 1 ) SecondHighestSalary;
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```
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## SQL Schema
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@ -756,7 +764,14 @@ VALUES
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CREATE FUNCTION getNthHighestSalary ( N INT ) RETURNS INT BEGIN
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SET N = N - 1;
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RETURN ( SELECT ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT N, 1 ) );
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RETURN (
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SELECT (
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SELECT DISTINCT Salary
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FROM Employee
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ORDER BY Salary DESC
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LIMIT N, 1
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)
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);
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END
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```
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@ -804,38 +819,93 @@ https://leetcode.com/problems/rank-scores/description/
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## Solution
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要统计某个 score 的排名，只要统计大于该 score 的 score 数量，然后加 1。
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要统计某个 score 的排名，只要统计大于等于该 score 的 score 数量。
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| score | 大于该 score 的 score 数量 | 排名 |
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| :---: | :---: | :---: |
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| 4.1 | 2 | 3 |
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| 4.2 | 1 | 2 |
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| 4.3 | 0 | 1 |
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| Id | score | 大于等于该 score 的 score 数量 | 排名 |
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| :---: | :---: | :---: | :---: |
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| 1 | 4.1 | 3 | 3 |
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| 2 | 4.2 | 2 | 2 |
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| 3 | 4.3 | 1 | 1 |
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但是在本题中，相同的 score 只算一个排名：
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使用连接操作找到某个 score 对应的大于其值的记录：
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| score | 排名 |
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```sql
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SELECT
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*
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FROM
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Scores S1
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INNER JOIN Scores S2
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ON S1.score <= S2.score
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ORDER BY
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S1.score DESC, S1.Id;
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```
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| S1.Id | S1.score | S2.Id | S2.score |
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| :---: | :---: | :---: | :---: |
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|3| 4.3| 3 |4.3|
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|2| 4.2| 2| 4.2|
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|2| 4.2 |3 |4.3|
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|1| 4.1 |1| 4.1|
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|1| 4.1 |2| 4.2|
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|1| 4.1 |3| 4.3|
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可以看到每个 S1.score 都有对应好几条记录，我们再进行分组，并统计每个分组的数量作为 'Rank'
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```sql
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SELECT
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S1.score 'Score',
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COUNT(*) 'Rank'
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FROM
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Scores S1
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INNER JOIN Scores S2
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ON S1.score <= S2.score
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GROUP BY
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S1.id, S1.score
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ORDER BY
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S1.score DESC, S1.Id;
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```
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| score | Rank |
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| :---: | :---: |
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| 4.1 | 3 |
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| 4.1 | 3 |
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| 4.2 | 2 |
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| 4.2 | 2 |
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| 4.3 | 1 |
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| 4.3 | 1 |
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| 4.2 | 2 |
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| 4.1 | 3 |
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可以按 score 进行分组，将同一个分组中的 score 只当成一个。
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上面的解法看似没问题，但是对于以下数据，它却得到了错误的结果：
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但是如果分组字段只有 score 的话，那么相同的 score 最后的结果只会有一个，例如上面的 6 个记录最后只取出 3 个。
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| score | 排名 |
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| Id | score |
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| :---: | :---: |
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| 4.1 | 3 |
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| 1 | 4.1 |
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| 2 | 4.2 |
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| 3 | 4.2 |
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| score | Rank |
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| :---: | :--: |
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| 4.2 | 2 |
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| 4.3 | 1 |
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| 4.2 | 2 |
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| 4.1 | 3 |
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所以在分组中需要加入 Id，每个记录显示一个结果。综上，需要使用 score 和 id 两个分组字段。
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而我们希望的结果为：
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在下面的实现中，首先将 Scores 表根据 score 字段进行自连接，得到一个新表，然后在新表上对 id 和 score 进行分组。
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| score | Rank |
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| :---: | :--: |
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| 4.2 | 1 |
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| 4.2 | 2 |
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| 4.1 | 2 |
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连接情况如下：
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| S1.Id | S1.score | S2.Id | S2.score |
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| :---: | :------: | :---: | :------: |
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| 2 | 4.2 | 3 | 4.2 |
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| 2 | 4.2 | 2 | 4.2 |
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| 3 | 4.2 | 3 | 4.2 |
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| 3 | 4.2 | 2 | 4.1 |
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| 1 | 4.1 | 3 | 4.2 |
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| 1 | 4.1 | 2 | 4.3 |
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| 1 | 4.1 | 1 | 4.1 |
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我们想要的结果是，把分数相同的放在同一个排名，并且相同分数只占一个位置，例如上面的分数，Id=2 和 Id=3 的记录都有相同的分数，并且最高，他们并列第一。而 Id=1 的记录应该排第二名，而不是第三名。所以在进行 COUNT 计数统计时，我们需要使用 COUNT( DISTINCT S2.score ) 从而只统计一次相同的分数。
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```sql
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SELECT
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@ -860,12 +930,12 @@ IF
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CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
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INSERT INTO Scores ( Id, Score )
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VALUES
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( 1, 3.5 ),
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( 2, 3.65 ),
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( 3, 4.0 ),
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( 4, 3.85 ),
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( 5, 4.0 ),
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( 6, 3.65 );
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( 1, 4.1 ),
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( 2, 4.1 ),
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( 3, 4.2 ),
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( 4, 4.2 ),
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( 5, 4.3 ),
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( 6, 4.3 );
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```
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# 180. Consecutive Numbers
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@ -585,7 +585,10 @@ SELECT
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FROM
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Customers
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WHERE
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Id NOT IN ( SELECT CustomerId FROM Orders );
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Id NOT IN (
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SELECT CustomerId
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FROM Orders
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);
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```
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## SQL Schema
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@ -666,7 +669,9 @@ SELECT
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FROM
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Employee E,
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Department D,
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( SELECT DepartmentId, MAX( Salary ) Salary FROM Employee GROUP BY DepartmentId ) M
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( SELECT DepartmentId, MAX( Salary ) Salary
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FROM Employee
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GROUP BY DepartmentId ) M
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WHERE
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E.DepartmentId = D.Id
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AND E.DepartmentId = M.DepartmentId
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@ -727,7 +732,10 @@ https://leetcode.com/problems/second-highest-salary/description/
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```sql
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SELECT
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( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT 1, 1 ) SecondHighestSalary;
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( SELECT DISTINCT Salary
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FROM Employee
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ORDER BY Salary DESC
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LIMIT 1, 1 ) SecondHighestSalary;
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```
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## SQL Schema
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@ -756,7 +764,14 @@ VALUES
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CREATE FUNCTION getNthHighestSalary ( N INT ) RETURNS INT BEGIN
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SET N = N - 1;
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RETURN ( SELECT ( SELECT DISTINCT Salary FROM Employee ORDER BY Salary DESC LIMIT N, 1 ) );
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RETURN (
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SELECT (
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SELECT DISTINCT Salary
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FROM Employee
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ORDER BY Salary DESC
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LIMIT N, 1
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)
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);
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END
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```
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@ -804,38 +819,93 @@ https://leetcode.com/problems/rank-scores/description/
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## Solution
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要统计某个 score 的排名，只要统计大于该 score 的 score 数量，然后加 1。
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要统计某个 score 的排名，只要统计大于等于该 score 的 score 数量。
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| score | 大于该 score 的 score 数量 | 排名 |
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| :---: | :---: | :---: |
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| 4.1 | 2 | 3 |
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| 4.2 | 1 | 2 |
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| 4.3 | 0 | 1 |
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| Id | score | 大于等于该 score 的 score 数量 | 排名 |
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| :---: | :---: | :---: | :---: |
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| 1 | 4.1 | 3 | 3 |
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| 2 | 4.2 | 2 | 2 |
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| 3 | 4.3 | 1 | 1 |
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但是在本题中，相同的 score 只算一个排名：
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使用连接操作找到某个 score 对应的大于其值的记录：
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| score | 排名 |
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```sql
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SELECT
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*
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FROM
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Scores S1
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INNER JOIN Scores S2
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ON S1.score <= S2.score
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ORDER BY
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S1.score DESC, S1.Id;
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```
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| S1.Id | S1.score | S2.Id | S2.score |
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| :---: | :---: | :---: | :---: |
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|3| 4.3| 3 |4.3|
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|2| 4.2| 2| 4.2|
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|2| 4.2 |3 |4.3|
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|1| 4.1 |1| 4.1|
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|1| 4.1 |2| 4.2|
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|1| 4.1 |3| 4.3|
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可以看到每个 S1.score 都有对应好几条记录，我们再进行分组，并统计每个分组的数量作为 'Rank'
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```sql
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SELECT
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S1.score 'Score',
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COUNT(*) 'Rank'
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FROM
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Scores S1
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INNER JOIN Scores S2
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ON S1.score <= S2.score
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GROUP BY
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S1.id, S1.score
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ORDER BY
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S1.score DESC, S1.Id;
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```
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| score | Rank |
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| :---: | :---: |
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| 4.1 | 3 |
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| 4.1 | 3 |
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| 4.2 | 2 |
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| 4.2 | 2 |
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| 4.3 | 1 |
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| 4.3 | 1 |
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| 4.2 | 2 |
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| 4.1 | 3 |
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可以按 score 进行分组，将同一个分组中的 score 只当成一个。
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上面的解法看似没问题，但是对于以下数据，它却得到了错误的结果：
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但是如果分组字段只有 score 的话，那么相同的 score 最后的结果只会有一个，例如上面的 6 个记录最后只取出 3 个。
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| score | 排名 |
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| Id | score |
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| :---: | :---: |
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| 4.1 | 3 |
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| 1 | 4.1 |
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| 2 | 4.2 |
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| 3 | 4.2 |
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| score | Rank |
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| :---: | :--: |
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| 4.2 | 2 |
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| 4.3 | 1 |
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| 4.2 | 2 |
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| 4.1 | 3 |
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所以在分组中需要加入 Id，每个记录显示一个结果。综上，需要使用 score 和 id 两个分组字段。
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而我们希望的结果为：
|
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|
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在下面的实现中，首先将 Scores 表根据 score 字段进行自连接，得到一个新表，然后在新表上对 id 和 score 进行分组。
|
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| score | Rank |
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| :---: | :--: |
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| 4.2 | 1 |
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| 4.2 | 2 |
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| 4.1 | 2 |
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连接情况如下：
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| S1.Id | S1.score | S2.Id | S2.score |
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| :---: | :------: | :---: | :------: |
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| 2 | 4.2 | 3 | 4.2 |
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| 2 | 4.2 | 2 | 4.2 |
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| 3 | 4.2 | 3 | 4.2 |
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| 3 | 4.2 | 2 | 4.1 |
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| 1 | 4.1 | 3 | 4.2 |
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| 1 | 4.1 | 2 | 4.3 |
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| 1 | 4.1 | 1 | 4.1 |
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我们想要的结果是，把分数相同的放在同一个排名，并且相同分数只占一个位置，例如上面的分数，Id=2 和 Id=3 的记录都有相同的分数，并且最高，他们并列第一。而 Id=1 的记录应该排第二名，而不是第三名。所以在进行 COUNT 计数统计时，我们需要使用 COUNT( DISTINCT S2.score ) 从而只统计一次相同的分数。
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|
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```sql
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SELECT
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@ -860,12 +930,12 @@ IF
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CREATE TABLE Scores ( Id INT, Score DECIMAL ( 3, 2 ) );
|
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INSERT INTO Scores ( Id, Score )
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VALUES
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( 1, 3.5 ),
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( 2, 3.65 ),
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( 3, 4.0 ),
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( 4, 3.85 ),
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( 5, 4.0 ),
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( 6, 3.65 );
|
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( 1, 4.1 ),
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( 2, 4.1 ),
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( 3, 4.2 ),
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( 4, 4.2 ),
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( 5, 4.3 ),
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( 6, 4.3 );
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```
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# 180. Consecutive Numbers
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