Class Practice (returns the number of the largest subarray in an integer array)

Source: Internet
Author: User
Tags sca

1. Design Ideas

It can be done with just one main function.

First, an integer array is defined, allowing the user to enter a set of integers.

Then, determine the number of positive and negative 0 of the situation.

Finally, choose the Execute if statement according to the judging situation, the case is all 0, negative number and 0, positive number and 0 positive negative number and 0 respectively.

2. Source code

`1 /*2 * For the maximum of the subarray of an integer array3 * The start time 2016/4/8 16:504 * The end of time 2016/4/8 18:305 * Author Jing6 */7  PackageArraymax;8 ImportJava.util.*;9  Public classSumarray {Ten  One      Public Static voidMain (string[] args) { A          -Scanner sca=NewScanner (system.in); -System.out.println ("Number of input integer arrays"); the         intnum=sca.nextint (); -          -         inta[]=New int[num],b[]=New int[num]; -         inti; +System.out.println ("Enter an array of integers for this group"); -          for(i=0;i<num;i++) +         { Aa[i]=sca.nextint (); at         } -         intL=0,j=0,k=0,sum=0, Max; -          for(i=0;i<num;i++)//judging the positive and negative conditions of an input array -         { -             if(a[i]>=0) -             { inJ + +; -             } to             if(a[i]<0) +             { -k++; the             } *             if(a[i]==0) \$             {Panax Notoginsengl++; -             } the         } +         if(K==num)//are all negative A         { theMax=a[0]; +              for(i=1;i<num;i++) -             { \$                 if(max<A[i]) \$                 { -max=A[i]; -                 } the             } -SYSTEM.OUT.PRINTLN ("Maximum sub-array and" +max);Wuyi         } the         Else if((l+k) ==num)//only negative numbers and 0 -         { WuSYSTEM.OUT.PRINTLN ("Maximum sub-array and 0"); -         } About         Else if(J==num)//are all non-negative \$         { -              for(i=0;i<num;i++) -             { -sum+=A[i]; A             } +SYSTEM.OUT.PRINTLN ("Maximum sub-array and" +sum); the         } -         Else \$         { the              for(i=0;i<num-1;i++) the             { the                 if(a[i]>=0&&a[i+1]>=0) the                 { -A[i+1]=a[i]+a[i+1]; inA[i]=0; the                 } the                 if(a[i]<0&&a[i+1]<0) About                 { theA[i+1]=a[i]+a[i+1]; theA[i]=0; the                 } +             } -K=0; theJ=0;Bayi              while(K<num)//only positive negative numbers in b[] after loop completion the             { the                 if(a[k]!=0) -                 { -b[j]=A[k]; theJ + +; the                 } thek++; the             } -             if(b[0]<0)//looking for the subscript of the first positive number the             { theI=1; the             }94             Else the             { theI=0; the             }98              while(i<j-2)//calculated only for positive and negative numbers About             { -                 if(B[i]> (-b[i+1]) && (-b[i+1]) <b[i+2])101                 {102B[i+2]=b[i]+b[i+1]+b[i+2];103B[i]=0;104B[i+1]=0; theI=i+2;106                 }107                 Else108                 {109I=i+2; the                 }111             } theMax=b[0];113              for(i=1;i<j;i++) the             { the                 if(max<B[i]) the                 {117max=B[i];118                 }119             } -SYSTEM.OUT.PRINTLN ("Maximum sub-array and" +max);121         }122     }123}`
The Main Code

3. Results

4. Programming Summary

Due to the time problem, this program is a bit hasty, there are many shortcomings, and need to optimize the place. In my spare time, I continued to think about the process of making up a minimum of my own satisfaction.

Class Practice (returns the number of the largest subarray in an integer array)

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