Find the contiguous subarray within an array (containing at least onenumber) which has the largest sum.

For example, given the array [?2,1,?3,4,?1,2,1,?5,4],
the contiguous subarray [4,?1,2,1] has thelargest sum = 6.

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 Divide and Conquer Array Dynamic Programming

#include<iostream>
using namespace std;

//取大值
int max(int a, int b, int c)
{
	if (a > b)
		if (a > c)
			return a;
		else
			return c;
	else if (c > b)
		return c;
	return b;
}

//flag=1，返回与A同尾的子数组的最大值，flag=0，返回与A同头的子数组的最大值
int maxSideArray(int *A, int n, int flag)
{
	if (flag)//返回与A同尾的子数组的最大值
	{
		int sum = 0;
		int maxsum = A[n - 1];
		for (int i = n - 1; i >= 0; i--)
		{
			sum += A[i];
			if (sum > maxsum)
				maxsum = sum;
		}
		return maxsum;
	}
	else//返回与A同头的子数组的最大值
	{
		int sum = 0;
		int maxsum = A[0];
		for (int i = 0; i < n; i++)
		{
			sum += A[i];
			if (sum > maxsum)
				maxsum = sum;
		}
		return maxsum;
	}

}

//法2：分而治之
int maxSubArray2(int A[], int n)
{
	if (n == 1)
		return *A;
	//左半部分最大值，有半部分最大值，跨过中线的最大值：三个值取最大
	return max(maxSubArray2(A, n / 2), maxSubArray2(A + n / 2, n - n / 2), maxSideArray(A, n / 2, 1) + maxSideArray(A + n / 2, n - n / 2, 0));
}

//法1：线性扫描
int maxSubArray(int A[], int n)
{
	int sum = 0;
	int maxsum = A[0];
	for (int i = 0; i < n; i++)
	{
		sum += A[i];
		if (sum > maxsum)
			maxsum = sum;
		if (sum < 0)
			sum = 0;
	}
	return maxsum;
}


void main()
{
	int A[] = { 1, -8, 6, 3, -1, 5, 7, -2, 0, 1 };
	int B[] = { -2, 1 };
	cout << maxSubArray(A, 10) << endl;
	cout << maxSubArray(B, 2) << endl; 
	cout << maxSubArray2(A, 10) << endl;
	cout << maxSubArray2(B, 2) << endl;
	system("pause");
}

53.Maximum Subarray（法1线性扫面法2分治法）