SQL挑战/难题:给定堆栈跟踪 - 如何在每个时间点找到顶部元素？

Question

• 我的真实用例是合并嵌套范围.我画了一些草图然后我看到范围开始和结束之间的相似性来叠加PUSH和POP操作.我明白解决这个问题也会解决原来的问题.

---

• 该运柱实际上可以从问题中移除.当val为NULL时,它是POP操作,否则它是PUSH操作.

---

谜题

表stack_trace包含以下列:

• i - 表示某个时间点的整数值.
• op - 2个可能的操作:I("in"/"push")和O("out"/"pop").

• val - "in"/"push"操作插入的值或"out"/"pop"操作的NULL.

  目标是在每个时间点(i)找到堆栈顶部的值.

例如

(NULL值在此表示为空格)

数据:

i   op  val 
--  --  --  
1   I   A   
2   I   B   
3   O
4   I   C
5   O    
6   O   

要求的结果:

i   top_of_stack_val
--  ----------------
1   A
2   B
3   A
4   C
5   A
6   

要求

• 解决方案应该是单个SQL查询(子查询很好).
• 只允许使用以下子句:SELECT,FROM,WHERE,GROUP BY,HAVING,ORDER BY.
• 不允许使用WITH子句(CTE - 公用表表达式).
• 不允许使用T-SQL,PL/SQL等.
• 不允许使用UDF(用户定义的函数).
• 不允许使用变量.

样本数据

create table stack_trace
(
    i       int
   ,op      char(1)
   ,val     char(1)
)
;

insert into stack_trace (i,op,val) values (1,'I','A');
insert into stack_trace (i,op,val) values (2,'I','B');
insert into stack_trace (i,op,val) values (3,'I','C');
insert into stack_trace (i,op,val) values (4,'I','D');
insert into stack_trace (i,op,val) values (5,'I','E');
insert into stack_trace (i,op)     values (6,'O');
insert into stack_trace (i,op)     values (7,'O');
insert into stack_trace (i,op)     values (8,'O');
insert into stack_trace (i,op,val) values (9,'I','F');
insert into stack_trace (i,op)     values (10,'O');
insert into stack_trace (i,op,val) values (11,'I','G');
insert into stack_trace (i,op,val) values (12,'I','H');
insert into stack_trace (i,op)     values (13,'O');
insert into stack_trace (i,op)     values (14,'O');
insert into stack_trace (i,op,val) values (15,'I','I');
insert into stack_trace (i,op,val) values (16,'I','J');
insert into stack_trace (i,op,val) values (17,'I','K');
insert into stack_trace (i,op,val) values (18,'I','L');
insert into stack_trace (i,op,val) values (19,'I','M');
insert into stack_trace (i,op)     values (20,'O');
insert into stack_trace (i,op,val) values (21,'I','N');
insert into stack_trace (i,op)     values (22,'O');
insert into stack_trace (i,op,val) values (23,'I','O');
insert into stack_trace (i,op)     values (24,'O');
insert into stack_trace (i,op,val) values (25,'I','P');
insert into stack_trace (i,op)     values (26,'O');
insert into stack_trace (i,op)     values (27,'O');
insert into stack_trace (i,op,val) values (28,'I','Q');
insert into stack_trace (i,op,val) values (29,'I','R');
insert into stack_trace (i,op)     values (30,'O');
insert into stack_trace (i,op)     values (31,'O');
insert into stack_trace (i,op)     values (32,'O');
insert into stack_trace (i,op)     values (33,'O');
insert into stack_trace (i,op)     values (34,'O');
insert into stack_trace (i,op)     values (35,'O');
insert into stack_trace (i,op,val) values (36,'I','S');
insert into stack_trace (i,op)     values (37,'O');
insert into stack_trace (i,op)     values (38,'O');
insert into stack_trace (i,op,val) values (39,'I','T');
insert into stack_trace (i,op,val) values (40,'I','U');
insert into stack_trace (i,op)     values (41,'O');
insert into stack_trace (i,op,val) values (42,'I','V');
insert into stack_trace (i,op,val) values (43,'I','W');
insert into stack_trace (i,op,val) values (44,'I','X');
insert into stack_trace (i,op)     values (45,'O');
insert into stack_trace (i,op)     values (46,'O');
insert into stack_trace (i,op,val) values (47,'I','Y');
insert into stack_trace (i,op)     values (48,'O');
insert into stack_trace (i,op)     values (49,'O');
insert into stack_trace (i,op,val) values (50,'I','Z');
insert into stack_trace (i,op)     values (51,'O');
insert into stack_trace (i,op)     values (52,'O');

要求的结果

i   top_of_stack_val
--  ----------------
1   A
2   B
3   C
4   D
5   E
6   D
7   C
8   B
9   F
10  B
11  G
12  H
13  G
14  B
15  I
16  J
17  K
18  L
19  M
20  L
21  N
22  L
23  O
24  L
25  P
26  L
27  K
28  Q
29  R
30  Q
31  K
32  J
33  I
34  B
35  A
36  S
37  A
38  
39  T
40  U
41  T
42  V
43  W
44  X
45  W
46  V
47  Y
48  V
49  T
50  Z
51  T
52  

Answer 1

这是一个很好的谜题.

由于我的主要DBMS是Teradata,我使用分析功能(需要TD14.10 +)为它编写了一个解决方案:

SELECT dt.*,
   -- find the last item in the stack with the same position
   Last_Value(val IGNORE NULLS)
   Over (PARTITION BY pos
         ORDER BY i) AS top_of_stack_val
FROM 
 ( 
   SELECT st.*,
      -- calculate the number of items in the stack
      Sum(CASE WHEN op = 'I' THEN 1 ELSE -1 end) 
      Over (ORDER BY i
            ROWS Unbounded Preceding) AS pos
   FROM stack_trace AS st
 ) AS dt;

此解决方案也适用于Oracle,但PostgreSQL和SQL Server不支持该IGNORE NULLS选项,LAST_VALUE并且模拟它非常复杂,例如,请参阅Itzk Ben-Gan的The Last non NULL Puzzle

编辑:事实上它并不复杂,我忘了Itzik的第二个解决方案,旧的背驮式技巧;-)

Martin Smith的方法适用于所有四个DBMS.