Question

我的任务是获取数字并打印它们的出现次数。我必须使用整数0来终止进程，但我使用字符串x进行处理。

这是我的代码：

import java.util.Scanner;

public class yasf{
    public static void main(String[] args) {
        int[] numbers = new int[101];
        int num;
        Scanner scan = new Scanner(System.in);
        System.out.println("Enter a number between 0-100 inclusive. x terminates:");
        String numString = scan.nextLine();

        // The input treminated with x;
        while (!numString.equalsIgnoreCase("x")) {
            num = Integer.parseInt(numString);
            numbers[num] = numbers[num] + 1;
            System.out.println("Enter a number between 0-100 inclusive. x terminates:");
            numString = scan.nextLine();
        }
        for (int i = 0; i <= 100; i++) {
        if (numbers[i] != 0 && numbers[i] == 1)
            System.out.println(i + ": " + numbers[i] + " time");
        else if (numbers[i] > 1 && numbers[i] != 0)
            System.out.println(i + ": " + numbers[i] + " times");
        else
            System.out.print("");
    }

}

}

我该怎么做？

Answer 1

如果您仍想接受“0”作为有效输入，请删除

的第一个实例

System.out.println("Enter a number between 0-100 inclusive. x terminates:");
String numString = scan.nextLine();

并使你的主要while循环看起来像这样

do{
  System.out.println("Enter a number between 0-100 inclusive. x terminates:");
  numString = scan.nextLine();
  num = Integer.parseInt(numString);
  numbers[num] = numbers[num]+1;
}while (!numString.equalsIgnoreCase("0"));

使用整数终止进程

1 个答案: