Maximum Contiguous Subsequence Sum of At Least Length L.

Given a sequence of n real numbers A(1) ... A(n), determine a contiguous subsequence A(i) ... A(j) of length atleast L for which the sum of elements in the subsequence is maximized.

Set sumL to the sum of the first L array elements.
Set runningSum = sumL.
Set maxSum = sumL
For k = L + 1 up to n, the length of the array:
Set sumL = sumL + A(k) - A(k - L)
Set runningSum = max(runningSum + A(k), sumL)
Set maxSum = max(maxSum, runningSum)
Output maxSum

public class MaxContSumL {
	public static void main(String[] args) {
		int[] A = { 0, 5, -3, -1, 2, -4, -1, 7, 8 };
		int L = 2;
		maxContiguousSumOfLenAtleastL(A, L);

		int[] B = { -5, -1, 2, -3, 0, -3, 3, };
		L = 3;
		maxContiguousSumOfLenAtleastL(B, L);
	}

	public static void maxContiguousSumOfLenAtleastL(int[] A, int L) {
		int len = A.length;
		int sumL = 0;
		for (int i = 0; i < L; i++) {
			sumL += A[i];
		}
		int runningSum = sumL;
		int maxSum = sumL;
		int start = 0, finish = 0;
		for (int i = L; i < len; i++) {
			sumL += (A[i] - A[i - L]);
			runningSum = Math.max(runningSum + A[i], sumL);
			
			if (runningSum > maxSum) {
				maxSum = runningSum;
				start = i - L + 1;
				finish = i;
			}
		}

		System.out.println("Max sum: " + maxSum);
		System.out.println("Indices: i=" + start + " : j=" + finish);
	}
}