The maxinum contiguous subsequence sum problem

(1)the obvious O(N^3) algorithm

/** *//** 
  cibic maximum contiguous subsequence sum algorithm.
 seqStart and seqEnd represent the actual best sequence
*/
  public static int maxSubsequenceSum(int [] a)
...{
   int maxSum =0;

      for (int i=0;i<a.length; i++)

        for (  int j =i; j<a.length ; j++)

       {

            int thisSum =0;

         for (int k= i; k<= j; k++)

            thisSum +=a[k];

           if(thisSum>maxSum)

            {

              maxSum  =thisSum;

             seqStart =i;

             seqEnd =j;

         }

     }
  
 return maxSum;
}

an improved O(N^2) algorithm

在O（N^3）基础上改进：除去了最内的循环原因在于重复的计算thisSum是不必要的

/**//*
    Quadratic maximum contigous subsequence sum algorithm
    seqStart and seqEnd represent the actual best sequence
*/
public static int maxSubsequenceSum (int [] a)
...{
      int maxSum = 0;
      for (int i=0;i<a.length;i++)
      ...{
          
          int thisSum=0;
          for (int j=i+1; j<a.length ; j++)
          ...{
              thisSum +=a[j];
              if(thisSum >maxSum)
              ...{
                  maxSum=thisSum;
                  seqStart = i;
                  seqEnd =j;
                  }
              }
          }
            return maxSum;
    }

a linear algorithm

Theorem : any i, let Ai..j be the first sequence ,with Si..j<0. Then ,for any i<=p<=j and

p<=q,Ap..q either is not a maximum contiguous subsequence or is equal to an

already seen maximum contiguous subsequence

证明略。

/**//*
    linear maximum contigous subsequence sum algorithm
    seqStart and seqEnd represent the actual best sequence
*/
public static int maxSubsequenceSum (int [] a)
...{
      int maxSum = 0;
      int thisSum =0;
      for (int i=0,j=0;j<a.length;j++)
      ...{
          
          thisSum+=a[i];
          
              if(thisSum >maxSum)
              ...{
                  maxSum=thisSum;
                  seqStart = i;
                  seqEnd =j;
              }else if(thisSum<0)...{
                  i=j+1;
                  thisSum=0;
                  }
          }
            return maxSum;
    }