基本上我必须先输入用户输入,直到输入0,然后找到数组中的最大数,负数和正数之和。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,因为数组中有一个未指定数量的元素。我一直认为这是错误,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
} //结束课
答案 0 :(得分:3)
问题是您将号码count
传入findMax
并且在尝试访问此行的上限时抛出ArrayIndexOutOfBoundsException
:
} else if (numbers[count] > max) {
记住Java中的数组是基于零的。此外,由于Java使用按值传递,您需要将结果分配给findMax
的输出:
max = findMax(numbers, count - 1, 0);
逻辑上,findMax
将始终返回 last 号码而不是最高号码。您需要从递归方法中返回临时max
:
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;
}
此版本在内部处理count
索引上限:
max = findMax(numbers, count, 0);