Maximum Subarray的通过率居然这么高，我提交了几次都是Wrong Answer，郁闷！

Find the contiguous subarray within an array (containing at least one number) 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 the largest sum = 6.

More practice:
If you have figured out the O(n) solution, try coding another solution using the divide and conquer approach, which is more subtle.

1， 最简单最容易想的办法是两重循环遍历， 不过显然将会是Time Limit Exceeded.

int maxSubArray(int A[], int n) 
    {
        int max = A[0];
        
        for(int i = 0; i < n; ++i)
        {
            int sum = A[i];
            for(int j = i + 1; j < n; ++j)
            {
                sum += A[j];
                if(sum > max)
                    max = sum;
            }
        }
        
        return max;
    }

2, 分治法

int recursiveMaxSubArray(int A[], int left, int right)
{
	if(left == right)return A[left];
	if(left == right-1)return max(max(A[left], A[right]), A[left]+A[right]);

	//divide
	int mid = (left + right)/2;
	int lmax = recursiveMaxSubArray(A, left, mid-1);
	int rmax = recursiveMaxSubArray(A, mid+1, right);

	//the max is [i..mid..j]
	int mmax = A[mid];

	for(int j = mid + 1, sum = mmax; j <= right; ++j)
	{
		sum += A[j];
		if(sum > mmax)
			mmax = sum;	
	}
	
	for(int j = mid - 1, sum = mmax; j >= left; --j)
	{
		sum += A[j];
		if(sum > mmax)
			mmax = sum;
	}

	return max(max(lmax, rmax), mmax);
}

int maxSubArray(int A[], int n) 
{
	return recursiveMaxSubArray(A, 0, n-1);
}

3, 算法 

Kadane‘s algorithm: containing at least one positive number
    
    Ref. 
    1, http://en.wikipedia.org/wiki/Maximum_subarray_problem
    2, http://blog.csdn.net/joylnwang/article/details/6859677.
    
    Lemma 1: Suppose A[i...j] is the max subarray, then sum(A[i...k]) >= 0, where i <=k < j.
    Proof. If sum(A[i...k]) < 0, then sum(A[k+1...j]) > sum(A[i...j]), contradiction.
       

 int maxSubArray(int A[], int n) 
    {
        int max = A[0], sum = 0;
        for(int i = 0; i < n; ++i)
        {
            sum += A[i];
            if(sum > max)
                max = sum;
            if(sum <= 0)
                sum = 0;
        }
        return max;
    }