国外版

算法题（国外版）/reformat-department-table.html

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+<p>Table: <code>Department</code></p>
+
+<pre>
++-------------+---------+
+| Column Name | Type    |
++-------------+---------+
+| id          | int     |
+| revenue     | int     |
+| month       | varchar |
++-------------+---------+
+(id, month) is the primary key of this table.
+The table has information about the revenue of each department per month.
+The month has values in [&quot;Jan&quot;,&quot;Feb&quot;,&quot;Mar&quot;,&quot;Apr&quot;,&quot;May&quot;,&quot;Jun&quot;,&quot;Jul&quot;,&quot;Aug&quot;,&quot;Sep&quot;,&quot;Oct&quot;,&quot;Nov&quot;,&quot;Dec&quot;].
+</pre>
+
+<p>&nbsp;</p>
+
+<p>Write an SQL query to reformat the table such that there is a department id column and a revenue column <strong>for each month</strong>.</p>
+
+<p>Return the result table in <strong>any order</strong>.</p>
+
+<p>The query result format is in the following example.</p>
+
+<p>&nbsp;</p>
+<p><strong>Example 1:</strong></p>
+
+<pre>
+<strong>Input:</strong> 
+Department table:
++------+---------+-------+
+| id   | revenue | month |
++------+---------+-------+
+| 1    | 8000    | Jan   |
+| 2    | 9000    | Jan   |
+| 3    | 10000   | Feb   |
+| 1    | 7000    | Feb   |
+| 1    | 6000    | Mar   |
++------+---------+-------+
+<strong>Output:</strong> 
++------+-------------+-------------+-------------+-----+-------------+
+| id   | Jan_Revenue | Feb_Revenue | Mar_Revenue | ... | Dec_Revenue |
++------+-------------+-------------+-------------+-----+-------------+
+| 1    | 8000        | 7000        | 6000        | ... | null        |
+| 2    | 9000        | null        | null        | ... | null        |
+| 3    | null        | 10000       | null        | ... | null        |
++------+-------------+-------------+-------------+-----+-------------+
+<strong>Explanation:</strong> The revenue from Apr to Dec is null.
+Note that the result table has 13 columns (1 for the department id + 12 for the months).
+</pre>