Maximum child columns and issues

Problem Description:GivenA sequence of k integers {N?1??,N?2??, ...,N?K?? }, "Continuous child column" is defined as {N?I??,N?I+1??, ...,< Span class= "Mord" >n ? j??  }, where  < Span class= "base textstyle uncramped" >1≤i≤j≤k. "Maximum child columns and" are defined as the and the largest of all contiguous child column elements. For example, given sequence {-2, 11,-4, 13,-5,-2}, its contiguous sub-columns {11,-4, 13} have the largest and 20. You are now asked to write a program that calculates the maximum sub-columns of a given integer sequence.

/span> 1. Method of exhaustive: Time complexity O (n^3)

Exhaustive method, calculated all possibilities, Time complexity O (n^3).

1 intMAXSUBSEQSUM1 (intA[],intsize)2 {3     intthissum,maxsum=0;4      for(inti =0; I < size;i++)5          for(intj = I;j < size;j++)6         {    7Thissum =0;8              for(intK = i;k <= j;k++)9Thissum + =A[k];Ten             if(Maxsum <thissum) OneMaxsum =thissum; A         } -     returnmaxsum; -}

2. Algorithm two: Time complexity O (n^2)

On the basis of the poor lifting method, slightly improved, less a for loop, the efficiency has improved, the code is easy to understand.

intMAXSUBSEQSUM2 (intA[],intsize) {    intthissum,maxsum=0;  for(inti =0; I < size;i++) {thissum=0;  for(intj = I;j < size;j++) {thissum+=A[j]; if(Maxsum <thissum) Maxsum=thissum; }        }    returnmaxsum;}

3. Divide and Conquer: Time Complexity O (NLOGN)

Adopt the idea of recursion. Maximum child column and there are three cases, the left largest child columns, the right largest child columns, across the middle of the largest sub-columns, take three of the largest. Recursion to the last namely, two numbers, comparing the left number of the right number or their and, fetching large.

1 //Divide and conquer2 intMAXSUBSEQSUM3 (intA[],intsize)3 {4     returnMaxsum (A,0, size-1);5 }6 7 intMaxsum (intA[],intLeftintRight )8 {9     if(left = =Right )Ten         returnA[left] >0? A[left]:0; One     intCenter = (left + right)/2;  A     intMaxleftsum =maxsum (a,left,center); -     intMaxrightsum = Maxsum (a,center+1, right); -     //find the left boundary maximum value the     intCenterto_left =0, Maxsum_centerto_left =0; -      for(inti = Center;i >= left;i--){ -Centerto_left + =A[i]; -         if(Maxsum_centerto_left <centerto_left) +Maxsum_centerto_left =Centerto_left; -     } +     //ask for the right boundary maximum value A     intCenterto_right =0, Maxsum_centerto_right =0; at      for(inti = center+1; I <= right;i++){ -Centerto_right + =A[i]; -         if(Maxsum_centerto_right <centerto_right) -Maxsum_centerto_right =Centerto_right; -     } -     returnMAXSUM3 (Maxleftsum,maxrightsum,maxsum_centerto_left +maxsum_centerto_right); in } - intMAXSUM3 (intAintBintc) to { +     return(A&GT;B?A:B) >c? (a>b?)a:b): C;  -}

4. Online Processing: Time complexity O (n)

The rule of online processing is to seek the child columns from the beginning and record the current maximum sub-columns, if the current sub-columns is negative, discard it. The code clearly illustrates this rule.

Personal understanding of online processing of the core: the current sub-column and if positive, it is always possible to make the following sub-column and become larger. If the current sub-columns is negative, it is not possible to be behind the child column and become larger, so discard.

1 //Online processing2 intMAXSUBSEQSUM4 (intA[],intsize)3 {4     intthissum=0, maxsum=0;5       for(inti =0; I < size;i++){6Thissum + =A[i];7         if(Maxsum <thissum)8Maxsum =thissum;9         Else if(Thissum <0)TenThissum =0; One     } A     returnmaxsum; -}

Organize lessons from Mooc Chen, He Chinming data structures

Maximum child columns and issues