从 n 个数中取连续的 m 个数, 使其和最大

穷举法的时间复杂度为 O(nm - m^2), 略去实现.

以下实现的时间复杂度为 O(n - m), 空间复杂度为 O(1).

public class Test {

    public static void main(String[] args) {
        int[] numberArr = {1, 2, 4, 6, 2, 1, 0, -1, 8, 10, 0};
        int range = 4;
        int result = find(numberArr, range);
        System.out.println("满足条件的子集的起始索引为: " + result);
    }

    public static int find(int[] numberArr, int range) {
        int target = 0, i = 0, j = range;
        int isum = 0, jsum = 0;

        for (; j < numberArr.length; i++, j++) {
            isum += numberArr[i];
            jsum += numberArr[j];
            if (isum < jsum) {
                target = j + 1 - range;
                isum = 0;
                jsum = 0;
            }
        }

        return target;
    }
}