Java Huawei Real Questions-Maximum Value of the Golden Treasure Chest

need:

   Alibaba, a impoverished woodcutter, accidentally discovered the treasure trove of a bandit group on his way to cut firewood. The treasure trove contained boxes numbered from 0 to N, and each box had a number on it.
Alibaba recites a spell number k (k<N), finds the maximum value of the sum of k consecutive treasure chest numbers, and outputs the maximum value.

Enter description
  Enter a string of numbers in the first line, separated by commas, for example: 2,10,-3,-8,40,5
  In the second line, enter the spell number, for example: 4. The size of the spell number is smaller than the number of treasure chests.

Output description
   maximum value

enter:

      2,10,-3,-8,40,5 ->numeric string

      4 ->Spell number size

Output:

      39

coding:

public class AiLibaba{
    public static void main(String[] args) {
        //1.列表数据
        Scanner scanner = new Scanner(System.in);
        System.out.print("输入一个数字字串：");
        String str = scanner.nextLine();
        //字符串数组
        String[] strNum = str.split(",");
        //集合对象
        List<Integer> list = new ArrayList<>();
        for (int i = 0; i < strNum.length; i++) {
            list.add(new Integer(strNum[i])); //添加到集合中
        }
        //2.求出最大值
        System.out.print("输入咒语数字:");
        int numK = scanner.nextInt();
        //调用方法
        int max = showMax(list, numK);
        System.out.println("最大值:" + max);
    }

    /**
     *
     * @param list 列表数据
     * @param numK  咒语数字
     * @return
     */
    private static int showMax(List<Integer> list, int numK) {
//        （1）定义两个指针，先求两个指针之间的和
        int left = 0;  //左边
        int right = numK - 1; //右边
        //累计和
        int sum = 0;
        for (int i = left; i <= right; i++) {
            sum += list.get(i); //累计
        }
        //最大值
        int max = sum;

//       （2） 两个指针右移，前面的和减掉移出去的（左侧），加上移入的（右侧）这样便得到下一个连续K区间的和。
        //右移
        while (right < list.size() - 1) {
            sum -= list.get(left++);
            sum += list.get(++right);
//        再用比较的方式，如果这个区间和比之前的大，那么就记录，否则继续右移
            if (sum > max)
                max = sum;
        }
        return max;
    }
}

Effect: