Pair development--the and of the maximal subarray of a ring-shaped one-dimensional array

Title: Returns the and of the largest sub-array in an integer array.
Requirements:
(1) Enter an array of shapes with positive and negative numbers in the array.
(2) One or more consecutive integers in the array make up a sub-array, each of which has a and.
(3) If the array a[0] ... A[j-1] next to each other, allowing a[i-1], ... A[n-1], a[0] ... A[J-1] and the largest.
(4) Returns the position of the maximum subarray at the same time.
(5) The maximum value of the and of all sub-arrays is evaluated. Requires a time complexity of O (n).
First, design ideas
The optimal solution to this problem must be the following two possibilities.
Perhaps one: the optimal solution does not span array[n-1] to array[0], which is the same as a non-annular array.
Possible two: The optimal solution spans array[n-1] to array[0], a new problem.
For the first case, we can be based on the non-circular array to seek, for MAX1, for the second case, the original problem can be converted to the smallest sub-array and the problem, and then the array of all elements and subtract the smallest sub-array and, then the result must be across a[n-1] to a[0] case the largest sub-array and, Set to Max2. The final result is the larger of the MAX1 and MAX2.
Example 1: There are arrays 5,-1,-6, 8, 2
When the max1=10,max2=15 is obtained, the larger max2 is taken as the result.
Example 2: There are arrays-6, 8, 1, 6, 1
When the max1=15,max2=14 is obtained, the larger max1 is taken as the result.

Second, the source program

1  PackageCom.java.lianxi;2 ImportJava.util.*;3  Public classLianxi3 {4      Public Static voidMain (string[] args)5     {6         intnum,i;7         intSum=0;8         intMax;9Scanner cin=NewScanner (system.in);TenSystem.out.print ("Please enter the length of the array:"); Onenum=cin.nextint (); A         intarray[]=New int[num]; -         intMax1=array[0]; -         intMin=array[0]; the          for(i=0;i<num;i++) -         { -array[i]=cin.nextint (); -         } +          for(i=0;i<num;i++) -         { +              if(sum<=0) A              { atsum=Array[i];  -              } -              Else -              { -sum=sum+Array[i]; -              } in              if(sum>max1) -              { tomax1=sum; +              } -         } theSYSTEM.OUT.PRINTLN ("Maximum value of the first case:" +max1); *          for(i=0;i<num;i++) $         {Panax Notoginseng              if(sum>=0) -              { thesum=Array[i]; +                  A              } the              Else +              { -sum=sum+Array[i]; $                 $              } -              if(sum<min) -              { themin=sum; -              }Wuyi         } the         intSum1=0; -          for(i=0;i<num;i++) Wu         { -sum1=sum1+Array[i]; About         } $         intMax2=sum1-min; -SYSTEM.OUT.PRINTLN ("Maximum value of the second case:" +max2); -         if(max1>max2) -         { Amax=max1; +         } the         Else -         { $max=Max2; the         } theSystem.out.println ("sub-array and maximum value:" +max); the     } the  -}

Third, the operation result:

Iv. Experience

The first is the design idea, at the beginning we in class and the students think of the same, is simply the sum of the largest sub-array, and then we write the process, the time complexity does not meet the requirements, a new way of thinking, that is, "sum-minimum sub-array and". The sum of the maximum number of words in both cases is the same as the previous non-circular array summation idea, the realization of the proportion of the simple array of gods and horses, and now has not implemented the return sub-array, ...

Pair development--the and of the maximal subarray of a ring-shaped one-dimensional array