Question

以下算法找到最大QAZ。元素的QAZ是在该元素的索引之后具有更高值的元素的数量。流行的答案是使用除法和解决使用O（nLog（n））..

数组未排序。在最坏的情况下，“Divide＆amp; Conquer”将触及所有n个元素。另外'for loop'会触及n / 2个元素更糟糕。它是如何n * log（n）..不应该是O（n ^ 2）..谢谢

 SELECT 
   siteid,
   linenum,
   worktype,
   COUNT(CASE WHEN status = 'APPR' THEN 1 ELSE NULL END)   AS [Approved],
   COUNT(CASE WHEN status = 'review' THEN 1 ELSE NULL END)   AS [Review],
   COUNT(CASE WHEN status = 'wmatl' THEN 1 ELSE NULL END)   AS [WaitMatl],
   COUNT(CASE WHEN status = 'comp' THEN 1 ELSE NULL END)   AS [Complete],
   COUNT(CASE WHEN status = 'incomp' THEN 1 ELSE NULL END)   AS [InComplete],
   COUNT(CASE WHEN status = 'closed' THEN 1 ELSE NULL END)   AS [Closed],
   COUNT(CASE WHEN status not in ('appr','wmatl') THEN 1 ELSE NULL END) AS [All_Completed],
   Count (*) as allrecords,
   /* The below divide by does not work */
   COUNT(CASE WHEN status not in ('appr','wmatl') THEN 1 ELSE null END)  / Count (*) as Completion_Ratio

  FROM workorder
  WHERE (siteid in ('p202','p203','p201')) AND (worktype in   ('mpm','ppm','tspm')) AND (istask ='0') 
AND (historyflag ='0')  AND (woclass = 'workorder') --AND (status not in ('comp','closed','review','incomp'))
AND (assetnum is not null) AND (maintby not in ('ms','ed')) 
AND targcompdate < DATEADD(mm,DATEDIFF(mm,0,GETDATE())-0,0)
GROUP BY siteid,linenum,worktype
Order by siteid, linenum,worktype

参考 - documentation

Answer 1

是的，在最坏的情况下，for循环触及n/2个元素。这里的重要事实是输入的划分尽可能均匀。算法在n元素上完成的工作量为O(T(n))，其中重复T由

定义

T(1) = O(1)
T(n) = 2 T(n/2) + O(n).

Master Theorem的案例2适用，因此T(n) = O(n log n)。这与合并排序非常相似。

对于未排序的数组，'分而治之'是否会产生O（log N）？

1 个答案: