Question

所以我试图计算这个程序中整数出现的次数。代码仍然不起作用，但我是否在正确的轨道上？

    public static void main(String[] args) 
    {
        Scanner scan = new Scanner(System.in);

        int userInput = 0;
        ArrayList<Integer> myList = new ArrayList<Integer>();
        int[] newArray = new int[myList.size()];
        int index1 = -1;
        int current;

        for (int num = 0; num <= userInput ; num++)
        {   
            System.out.println("Please enter a random number between 0 and 50, enter a negative number to end input: ");
            num--;

            if(userInput >= 0 || userInput <= 50)
            {       
                userInput++;
                userInput = scan.nextInt();
                index1++;
                myList.add(userInput);

            }
            if (userInput < 0 || userInput > 50)
            {
                myList.remove(index1);
                index1--;
                break;
            }
        }   

        for (int num2 = 0; num2 < myList.size(); num2++)
            {
                current = myList.get(num2);
                if(current == myList.get(num2))
                {
                    newArray[current-myList.get(num2)]++;
                }
            }

            for (int number=0; number < newArray.length; number++)
            {
                System.out.println(number + "1");
                System.out.println(" : " + newArray[number]);
            }
}
}

编辑：只想添加我可以运行程序，但是当我输入一个超出范围的整数（不在0到50之间）时，我收到错误

Answer 1

我刚刚写完这篇文章，感谢我在修改我的代码时的downvotes，并特别声明我正在修改代码。 Hashmaps会在这里为你做到这一点。

public static void main(String args[]) {
        HashMap<Integer, Integer> numbers = new HashMap<Integer, Integer>();
        Scanner scan = new Scanner(System.in);
        boolean loop = true;
        while (loop) {
            System.out.println("Number: ");
            int current = scan.nextInt();
            if (current > 0 && current < 51) {
                if (numbers.containsKey(current)) {
                    numbers.put(current, numbers.get(current)+1);
                }else{
                    numbers.put(current, 1);
                }
            } else {
                if(numbers.isEmpty()){
                    System.out.println("Nothing inputted");
                    loop = false;
                }else{
                    for(Entry<Integer,Integer> e : numbers.entrySet()){
                        System.out.println(e.getKey() + " was entered " + e.getValue() + " times");
                    }
                    loop = false;
                }
            }
        }
    }

Answer 2

您不需要像myList这样的任何其他集合。你可以声明数组

int[] occurrences = new int[51];

可以存储范围[0,50]中元素的出现。当您从扫描仪读取 n 作为nextInt时，如果 n 在范围内或者停止读取，则应该在readed n 位置增加此数组的元素否则就是新的价值观。

如果statemest总是如此：

current = myList.get(num2);
if(current == myList.get(num2))

代码示例：

public static void main(String[] args) {

        Scanner scanner = new Scanner(System.in);
        int[] occurrrences = new int[51];

        while (true){
            int value = scanner.nextInt();
            if(value>=0 && value<occurrrences.length){
                occurrrences[value]++;
            }else{
                scanner.close();
                break;
            }
        }
        for (int i = 0; i < occurrrences.length; i++) {
            if(occurrrences[i]>0){
                System.out.println(i + " occured " + occurrrences[i] + " times.");
            }
        }
    }

试图计算数组

2 个答案: