Oracle time complexity的MAX函数O（1），O（log n）或O（n）是否与表中的行数有关？
答案 0 :(得分：9)
如果列上有B树索引，则查找最大值为O（log（n）），因为答案将是索引的最后一行（或第一行）。值存储在B树的最深节点中，其具有高度O（log（n））。
没有索引，它是O（n），因为必须读取所有行以确定最大值。
注意：O（n）表示法忽略常量，但在现实世界中，这些常量不能被忽略。从磁盘读取和从内存读取之间的差异是几个数量级。访问索引的第一个值可能主要在RAM中执行，而大表的全表扫描则需要主要从磁盘读取。
答案 1 :(得分：7)
实际上，如果不指定查询，表定义和查询计划，很难说。
如果您在列上没有索引的表中计算MAX，则Oracle必须执行全表扫描。那将是O（n）因为您必须扫描表中的每个块。您可以通过查看查询计划来查看。
我们将生成一个包含100,000行的表，并使用CHAR(1000)列确保行数相当大
SQL> create table foo( col1 number, col2 char(1000) );
Table created.
SQL> insert into foo
2 select level, lpad('a',1000)
3 from dual
4 connect by level <= 100000;
100000 rows created.
现在，我们可以查看基本MAX操作的计划。这是在进行全表扫描（O（n）操作）
SQL> set autotrace on;
SQL> select max(col1)
2 from foo;
MAX(COL1)
----------
100000
Execution Plan
----------------------------------------------------------
Plan hash value: 1342139204
---------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
---------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 13 | 4127 (1)| 00:00:50 |
| 1 | SORT AGGREGATE | | 1 | 13 | | |
| 2 | TABLE ACCESS FULL| FOO | 106K| 1350K| 4127 (1)| 00:00:50 |
---------------------------------------------------------------------------
Note
-----
- dynamic sampling used for this statement (level=2)
Statistics
----------------------------------------------------------
29 recursive calls
1 db block gets
14686 consistent gets
0 physical reads
176 redo size
527 bytes sent via SQL*Net to client
523 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
如果您在列上计算MAX的索引，则Oracle可以对索引执行MIN/MAX scan。如果这是优化程序选择的计划，那么这是一个O（log n）操作。当然，作为一个实际问题，这在功能上是一个O（1）操作，因为索引的高度实际上永远不会超过4或5--这里的常数项将占主导地位。
SQL> create index idx_foo_col1
2 on foo( col1 );
Index created.
SQL> select max(col1)
2 from foo;
MAX(COL1)
----------
100000
Execution Plan
----------------------------------------------------------
Plan hash value: 817909383
-------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
-------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 13 | 2 (0)| 00:00:01 |
| 1 | SORT AGGREGATE | | 1 | 13 | | |
| 2 | INDEX FULL SCAN (MIN/MAX)| IDX_FOO_COL1 | 1 | 13 | 2 (0)| 00:00:01 |
-------------------------------------------------------------------------------------------
Note
-----
- dynamic sampling used for this statement (level=2)
Statistics
----------------------------------------------------------
5 recursive calls
0 db block gets
83 consistent gets
1 physical reads
0 redo size
527 bytes sent via SQL*Net to client
523 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
但事情变得更难了。 MIN和MAX都具有相同的O（log n）行为。但是，如果同一查询中同时包含MIN和MAX，则会突然返回O（n）操作。 Oracle（截至11.2）尚未实现选项同时获取索引的第一个块和最后一个块
SQL> ed
Wrote file afiedt.buf
1 select min(col1), max(col1)
2* from foo
SQL> /
MIN(COL1) MAX(COL1)
---------- ----------
1 100000
Execution Plan
----------------------------------------------------------
Plan hash value: 1342139204
---------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
---------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | 13 | 4127 (1)| 00:00:50 |
| 1 | SORT AGGREGATE | | 1 | 13 | | |
| 2 | TABLE ACCESS FULL| FOO | 106K| 1350K| 4127 (1)| 00:00:50 |
---------------------------------------------------------------------------
Note
-----
- dynamic sampling used for this statement (level=2)
Statistics
----------------------------------------------------------
4 recursive calls
0 db block gets
14542 consistent gets
0 physical reads
0 redo size
601 bytes sent via SQL*Net to client
523 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed
当然，在后续版本的Oracle中，可能会实现此优化，这将回到O（log n）操作。当然，您也可以重写查询以获得不同的查询计划，该计划可以回到O（log n）
SQL> ed
Wrote file afiedt.buf
1 select (select min(col1) from foo) min,
2 (select max(col1) from foo) max
3* from dual
SQL>
SQL> /
MIN MAX
---------- ----------
1 100000
Execution Plan
----------------------------------------------------------
Plan hash value: 3561244922
-------------------------------------------------------------------------------------------
| Id | Operation | Name | Rows | Bytes | Cost (%CPU)| Time |
-------------------------------------------------------------------------------------------
| 0 | SELECT STATEMENT | | 1 | | 2 (0)| 00:00:01 |
| 1 | SORT AGGREGATE | | 1 | 13 | | |
| 2 | INDEX FULL SCAN (MIN/MAX)| IDX_FOO_COL1 | 1 | 13 | 2 (0)| 00:00:01 |
| 3 | SORT AGGREGATE | | 1 | 13 | | |
| 4 | INDEX FULL SCAN (MIN/MAX)| IDX_FOO_COL1 | 1 | 13 | 2 (0)| 00:00:01 |
| 5 | FAST DUAL | | 1 | | 2 (0)| 00:00:01 |
-------------------------------------------------------------------------------------------
Note
-----
- dynamic sampling used for this statement (level=2)
Statistics
----------------------------------------------------------
7 recursive calls
0 db block gets
166 consistent gets
0 physical reads
0 redo size
589 bytes sent via SQL*Net to client
523 bytes received via SQL*Net from client
2 SQL*Net roundtrips to/from client
0 sorts (memory)
0 sorts (disk)
1 rows processed