求解最大子序列和的经典实现

                                                      求解最大子序列和

        记录下最大序列和的多个实现方法，时间复杂度由高至低，分别为ON3、ON2、ONlogN、ON。分别对应的是：直接穷举式、穷举改进式、分治处理、联机算法。好的算法实现，总给人以美的体验。

        下面直接贴代码：

#include 

int MaxSubsequenceSum1(const int *PArr,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 += PArr[k];
            }
            if(ThisSum > MaxSum)
            {
                MaxSum= ThisSum;
            }
        }
    }
    return MaxSum;
    
}

int MaxSubsequenceSum2(const int *PArr,int n)
{
	int ThisSum, MaxSum, i, j;
	
	MaxSum=0;
	for(i=0; i< n; i++)
	{
		ThisSum= 0;
		for(j= i;j< n; j++)
		{
			ThisSum += PArr[j];
			
			if(ThisSum > MaxSum)
			{
				MaxSum= ThisSum;
			}
		}
	}
	return MaxSum;
}

int Max3(int a, int b, int c)
{
	return (a >= b)?((a >= c)?(a):(c)):((b >= c)?(b):(c));
}
static int MaxSubSum(const int *PArr,int Left,int Right)
{
	int MaxLeftSum, MaxRightSum;
	int MaxLeftBorderSum, MaxRightBorderSum;
	int LeftBorderSum, RightBorderSum;
	int Center, i;

	if(Left == Right)
		if(PArr[Left] > 0)
			return PArr[Left];
		else
			return 0;

	
	Center= (Left + Right)/2;
	MaxLeftSum= MaxSubSum(PArr, Left, Center);
	MaxRightSum= MaxSubSum(PArr, Center+1, Right);
	
	MaxLeftBorderSum= 0;
	LeftBorderSum= 0;
	for(i = Center; i >= Left; i--)
	{
		LeftBorderSum += PArr[i];
		if(LeftBorderSum > MaxLeftBorderSum)
			MaxLeftBorderSum= LeftBorderSum;
	}
	
	MaxRightBorderSum= 0;
	RightBorderSum= 0;
	for(i= Center + 1; i <= Right; i++)
	{
		RightBorderSum += PArr[i];
		if(RightBorderSum > MaxRightBorderSum)
			MaxRightBorderSum= RightBorderSum;
	}
	return Max3(MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum);
}

int MaxSubsequenceSum3(const int *PArr,int n)
{
	return MaxSubSum(PArr, 0, n-1);
}


int MaxSubsequenceSum4(const int *PArr,int n)
{
	int ThisSum, MaxSum, j;
	
	ThisSum= MaxSum= 0;
	for(j= 0; j< n; j++)
	{
		ThisSum += PArr[j];
		
		if(ThisSum > MaxSum)
			MaxSum= ThisSum;
		else if(ThisSum < 0)
			ThisSum= 0;
	}
	return MaxSum;
}
int main(void) { 
    int testArr[10]= {1,3,5,7,-9,-5,4,8,10,0};
	printf("%d\n",MaxSubsequenceSum1(testArr,10));
	printf("%d\n",MaxSubsequenceSum2(testArr,10));
	printf("%d\n",MaxSubsequenceSum3(testArr,10));
	printf("%d\n",MaxSubsequenceSum4(testArr,10));
	return 0;
}