Question

我在接受采访时得到以下问题：给定一个具有一些缺失的自然数的表，提供两个表的输出，在第一个表中的数字间隙的开始并以秒结束。例如：

 ____    ________
|    |   |   |   |
| 1  |   | 3 | 3 |
| 2  |   | 6 | 7 |
| 4  |   | 10| 12|
| 5  |   |___|___|
| 8  |
| 9  |
| 13 |
|____|

Answer 1

虽然这与Phil Sandler的答案几乎相同，但这应该返回两个单独的表（我觉得它看起来更干净）（至少在SQL Server中有效）：

DECLARE @temp TABLE (num int)
INSERT INTO @temp VALUES (1),(2),(4),(5),(8),(9),(13)

DECLARE @min INT, @max INT
SELECT @min = MIN(num), @max = MAX(num) FROM @temp

SELECT t.num + 1 AS range_start
    FROM @temp t
    LEFT JOIN @temp t2 ON t.num + 1 = t2.num
    WHERE t.num < @max AND t2.num IS NULL

SELECT t.num - 1 AS range_end
    FROM @temp t
    LEFT JOIN @temp t2 ON t.num - 1 = t2.num
    WHERE t.num > @min AND t2.num IS NULL

Answer 2

Itzik Ben Gan写了很多关于这些“差距和岛屿”问题的文章。他的row_number解决方案是

WITH C AS
(
SELECT N, ROW_NUMBER() OVER (ORDER BY N) AS RN
FROM t
)
SELECT Cur.N+1,Nxt.N-1
FROM C AS Cur 
JOIN C AS Nxt ON Nxt.RN = Cur.RN+1
WHERE Nxt.N-Cur.N>1

来自同一来源的没有row_number的解决方案。

SELECT N+1 AS start_range,
(SELECT MIN(B.N) FROM t AS B WHERE B.N > A.N)-1 AS end_range
FROM t AS A
WHERE NOT EXISTS(SELECT * FROM t AS B WHERE B.N = A.N+1)
AND N< (SELECT MAX(N) FROM t)

Answer 3

这可以在没有DB特定SQL的情况下工作，它可能会变得更干净但它确实有效

修改您可以在StackExchange Data Explorer 上看到这个Query

SELECT low,high FROM 

(

SELECT col1, low 

FROM
(Select n1.col1 col1, min(n2.col1) + 1 low
 from numbers n1
inner join numbers n2
on n1.col1 < n2.col1 

Group by n1.col1) t
WHERE t.low not in (SELECT col1 FROM NUMBERS)
and t.low < (Select MAX(col1) from numbers) 
) t

INNER JOIN 
(

SELECT col1 - 1 col1, high
 FROM
(Select n1.col1 col1 , min(n2.col1) - 1 high
 from numbers n1
inner join numbers n2
on n1.col1 < n2.col1 

Group by n1.col1) t
WHERE t.high not in (SELECT col1 FROM NUMBERS) 
) t2
ON t.col1 = t2.col1

Answer 4

这样的事情：

SELECT col1, col2 FROM
(
    SELECT x + 1 as col1, 
        ROW_NUMBER() OVER(ORDER BY x) AS 'rownum'  
    FROM tbl y 
    WHERE NOT EXISTS (SELECT x FROM tbl z WHERE z.x = y.x + 1) 
        AND x <> (SELECT MAX(x) FROM tbl)
) a
INNER JOIN
(
    SELECT x - 1 as col2,
        ROW_NUMBER() OVER(ORDER BY x) AS 'rownum'  
    FROM tbl y 
    WHERE NOT EXISTS (SELECT x FROM tbl z WHERE z.x = y.x - 1) 
        AND x <> (SELECT MIN(x) FROM tbl)
) b
ON a.rownum = b.rownum

对于不同的DBMS，“rownum”语法将有所不同。以上内容可能适用于SQL Server，但我还没有测试过它。

正如其中一条评论所指出的那样，许多DBMS都有分析功能，可以让这更容易。

Answer 5

这是SQL Server语法：

CREATE TABLE #temp (columnA int)

INSERT INTO #temp VALUES(1)
INSERT INTO #temp VALUES(2)
INSERT INTO #temp VALUES(4)
INSERT INTO #temp VALUES(5)
INSERT INTO #temp VALUES(8)
INSERT INTO #temp VALUES(9)
INSERT INTO #temp VALUES(13)

SELECT 
    t1.columnA - 1
FROM 
    #temp t1
    LEFT JOIN #temp t2 ON t1.columnA = t2.ColumnA + 1
WHERE 
    t2.ColumnA IS NULL
    AND t1.ColumnA != (SELECT MIN(ColumnA) from #temp)  

SELECT 
    t1.columnA + 1
FROM 
    #temp t1
    LEFT JOIN #temp t2 ON t1.columnA = t2.ColumnA - 1
WHERE 
    t2.ColumnA IS NULL  
    AND t1.ColumnA != (SELECT MAX(ColumnA) from #temp)  

DROP table #temp

Answer 6

您可以使用Lag功能访问上一行：

create table #a (n int)

insert #a values(1)
insert #a values(2)
insert #a values(4)
insert #a values(5)
insert #a values(8)
insert #a values(9)
insert #a values(13)

select  prev + 1, n - 1 from
(select lag(n) over(order by n) as prev, n
from    #a) a
where   prev < n - 1

结果：

|3  |3  |

|6  |7  |

|10 |12 |

Answer 7

SQL Fiddle Setup and Solution

1。第1步

从ID列表中获取当前ID和所有可用的下一个ID

select l1.id curr_id,l2.id next_id from
id_list l1,id_List l2
where l1.id < l2.id;

2。步骤2

从上面的列表中，我们将看到所有组合，但每个当前ID仅过滤一个组合，其下一个ID最小，为此，获取每个当前ID的最小当前ID和最小下一个ID。按当前ID使用分组方式

with id_combinations as
(
 select l1.id curr_id,l2.id next_id from
 id_list l1,id_List l2
 where l1.id < l2.id
)
select min(curr_id)+1 missing_id_start -- Need to add 1 from current available id
       ,min(next_id)-1 missing_id_end -- Need to subtract 1 from next available id 
from id_combinations
group by curr_id
having min(curr_id)+1 < min(next_id) -- Filter to get only the missing ranges

SQL面试问题

7 个答案: