最大子序列和问题的四种算法

算法一,该算法的复杂度为：O(N^3)

int MaxSubsequenceSum( const int A[],int N)
{
	int ThisSum,Maxsum,i,j,k;
    MaxSum = 0;
    for(i = 0; i < N; i++)
    	for(j = i; j < N; j++)
        {
        	ThisSum = 0;
            for(k = i; k <= j; k++)
            	ThisSum += A[k];
            if(ThisSum > MaxSum)
            	MaxSum = ThisSum;
        }
    return MaxSum;
}

算法二，该算法的复杂度为：O(N^2)

int MaxSubsequenceSum( const int A[],int N)
{
	int ThisSum,MaxSum,i,j;
    MaxSum = 0;
    for( i = 0; i < N; i++)
    {
    	ThisSum = 0;
        for(j = i; j < N;j++)
        {
        	ThisSum += A[j];
            if(ThisSum > MaxSum)
            	MaxSum = ThisSum;
        }
    }
    return MaxSum;
}

算法三，该算法的复杂度为：O(NlogN)

static int MaxSubSum(const int A[],int Left,int Right)
{
	int MaxLeftSum,MaxRightSum;
    int MaxLeftBorderSum,MaxRightBorderSum;
    int LeftBorderSum,RightBorderSum;
    int Center,i;
    
    if (Left == Right) /* Base Case */
    	if (A[Left] > 0)
        	return A[Left];
        else
        	return 0;
    Center = (Left + Right)/2;
    MaxLeftSum = MaxSubSum(A,Left,Center);
    MaxRightSum = MaxSubSum(A,Center+1,Right);
   
    MaxLeftBorderSum = 0;
    LeftBorderSum = 0;
    for(i = Center; i >= Left; i--)
    {
   	   LeftBorderSum += A[i];
          if(LeftBorderSum > MaxLeftBorderSum)
        	MaxLeftBorderSum = LeftBorderSum;
    }
   
    MaxRightBorderSum = 0;
    RightBorderSum = 0;
    for(i = Center + 1; i <= Right; i++)
    {
   	  RightBorderSum += A[i];
         if(RightBorderSum > MaxRightBorderSum)
        	MaxRightBorderSum = RightBorderSum;
   }
   
   return Max3(MaxLeftSum,MaxRightSum,MaxLeftBorderSum + MaxRightBorderSum);
}
int MaxSubsequenceSum(const int A[],int N)
{
	return MaxSubSum(A,0,N-1);
}

算法四，该算法的复杂度为：O(N)

int MaxSubsequenceSum(const int A[],int N)
{
	int ThisSum,MaxSum,j;
    ThisSum = MaxSum = 0;
    for(j =0; j < N; j++)
    {
    	ThisSum += A[j];
        if(ThisSum > MaxSum)
        	MaxSum = ThisSum;
        else if(ThisSum < 0)
        	ThisSum = 0;
    }
    return MaxSum;
}