求解最大子序列和的四種方法

1、三重循環暴力求解

例程：

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

分析復雜度：

復雜度為

\[\sum^{n}_{i = 1}(\sum_{j = i}^n(j - i)) \]

而

\[\sum_{j = i}^n(j - i) = \frac{(n - i)(n - i + 1)}{2} = \frac{(n - i)^2}{2} + \frac{n- i}{2} \]

\(\Rightarrow\)

\[\sum_{i = 1}^n(\frac{(n - i)^2}{2} + \frac{n - i}{2}) = O(n^3) \]

2、兩重循環

上面的三重循環可以撤除一個，以此來獲取平方級別的復雜度

例程：

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

易知，復雜度為

\[\sum_{i = 1}^{n}(n - i) = O(n^2) \]

3、分治

例程：

static int MaxSubSum(const int A[], int Left, int Right)
{
	int MaxLeftSum, MaxRightSum; // 左右部分各自的最大子序列和
	int MaxLeftBorderSum, MaxRightBorderSum; // 包括邊界的左右最大子序列和的最大值(從中間開始往左右延伸)，這兩個數最后要相加到一起，然后和上面的兩個數比較，它們的和的本質其實是跨越中間邊界的最大子序列和
	int LeftBorderSum, RightBorderSum; // 存儲臨時值
	int center, i;
	
	if(Left == Right)
		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 > MaxRightBorerSum)
			MaxRightBorderSum = RightBorderSum;
	}
	
	return Max3(MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum);
}

int MaxSubSequenceSum(const int A[], int N)
{
	return MaxSubSum(A, 0, N - 1);
}

\[T(n) = 2T(\frac{n}{2}) + O(n) \]

\(\Rightarrow\)

\[T(n) = O(nlogn) \]

4、聯機算法（online algorithm）

見另一篇文章：https://www.cnblogs.com/fanlumaster/p/13654938.html