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基本上我必须接受用户输入，直到他们输入 0，然后找到数组中的最大值、负数的数量和正数的总和。问题是我必须使用和数组并且不能使用 ArrayList 来调用其他方法，所以我想我会创建一个 ArrayList 并使用它的元素来创建一个数组，因为数组中有未指定数量的元素。我一直把它当作错误，并尝试了我能想到的一切。请帮忙。

Exception in thread "main" java.lang.IndexOutOfBoundsException: Index: 3, Size: 3
    at java.util.ArrayList.RangeCheck(ArrayList.java:547)
    at java.util.ArrayList.get(ArrayList.java:322)
    at Assignment9.assignment9(Assignment9.java:37)
    at Assignment9.assignment9(Assignment9.java:49)
    at Assignment9.assignment9(Assignment9.java:49)
    at Assignment9.assignment9(Assignment9.java:49)
    at Assignment9.main(Assignment9.java:17)

//Class Description: Takes user input and makes it into an array until 0 is entered.
// It then computes the maximum #, amount of negative #s and the sum of the positive #s.
import java.util.*;
import java.text.*;;

public class Assignment9 {

//main method initializes variables then calls method assignment9
public static void main(String[] args)
{
    int count = 0;
    ArrayList<Double> help = new ArrayList<Double>();
    assignment9(help, count);
}//end main

//adds user input to array until 0 is entered; it then calls other methods
public static void assignment9(ArrayList<Double> help, int count)
{
    DecimalFormat fmt = new DecimalFormat("#.##");
    double input;
    double max = 0;
    int negative = 0;
    double sum = 0;
    Scanner scan = new Scanner(System.in);

    input = scan.nextInt();

    if (input == 0)
    {
        double[] numbers = new double[help.size()];
        for(int i = 0; i < numbers.length; i++)
        {
            numbers[i] = help.get(i);
        }
        findMax(numbers, count, max);
        countNegative(numbers, count, negative);
        computeSumPositive(numbers, count, sum);
        System.out.println("The maximum numer is " + fmt.format(max));
        System.out.println("The total number of negative numbers is " + negative);
        System.out.println("The sum of positive numbers is " + fmt.format(sum));
        System.exit(0);
    }else{
            help.add(input);
            count++;
            assignment9(help, count);
    }//end if
}//end assignment9

//compares elements of array to find the max until count = -1
public static double findMax(double[] numbers, int count, double max)
{

    if (count == -1)
    {
        return max;
    }else if(numbers[count] > max){
        max = numbers[count];
        count--;
        findMax(numbers, count, max);
    }//end if
    return max;

}//end findMax

public static int countNegative(double[] numbers, int count, int negative)
{

    if(count == -1)
    {
        return negative;
    }else if(numbers[count] < 0){
        negative++;
        count--;
        countNegative(numbers, count, negative);
    }
    return negative;
}//end countNegative

public static double computeSumPositive(double[] numbers, int count, double sum)
{
     if(count == -1)
     {
         return sum;
     }else{
         sum = sum + numbers[count];
         count++;
         computeSumPositive(numbers, count, sum);
         return sum;
     }
}//end computeSumPositive

}//结束类

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1 回答 1

3

问题是您正在将数字传递count给并且尝试访问此行的上限时findMax会抛出一段时间：ArrayIndexOutOfBoundsException

} else if (numbers[count] > max) {

请记住，Java 中的数组是从零开始的。此外，由于 Java 使用按值传递，您需要将结果分配给以下输出findMax：

max = findMax(numbers, count - 1, 0);

从逻辑上讲，findMax将始终返回最后一个数字而不是最高数字。max您需要从递归方法中返回临时值：

double findMax(double numbers[], int count, int index) {
   // This needs to be size - 1 because the array is 0-indexed.
   // This is our base case
   if (index == count - 1)
      return numbers[index];

   // Call the function recursively on less of the array
   double result = findMax(numbers, count, index + 1);

   // Return the max of (the first element we are examining, the max of the
   // rest of the array)
   if (numbers[index] > result)
      return numbers[index];
   else
      return result;
}

此版本在count内部处理索引上限：

max = findMax(numbers, count, 0);
于 2013-03-29T21:23:22.710 回答