Question

import java.util.Scanner;

public class TomAndJerry {
    Scanner sc = new Scanner(System.in);

    public String catAndMouse(int x, int y, int z) {
        String result = null;
        int q = 0;
        System.out.println("enter the number of queries");
        q = sc.nextInt();
        for (int i = 0; i < q; i++) {
            x = sc.nextInt();
            y = sc.nextInt();
            z = sc.nextInt();
        }

        if (Math.abs(x - z) < Math.abs(y - z)) {
            result = "Cat A";
        } else if (Math.abs(x - z) > Math.abs(y - z)) {
            result = "Cat B";
        } else {
            result = "Mouse C";
        }
        return result;
    }

    public static void main(String[] args) {
        int x = 0, y = 0, z = 0;
        String result = null;
        TomAndJerry tj = new TomAndJerry();
        result = tj.catAndMouse(x, y, z);
        System.out.println(result);
    }

}

输出：-

 enter the number of queries
                    2
                  1 2 3
                  1 3 2
                 Mouse C

我最终只获得使用扫描仪提供的最后一个查询/输入的输出。我如何获得其余查询/输入的输出。谢谢您的宝贵时间！：）

Answer 1

您需要使用List或String []而不是简单的String对象才能从一个函数中返回多个String，同样，if条件需要成为for循环的一部分，并且需要将结果添加到List或数组中，下面是使用List的预期输出

import java.util.*;

public class Main {
Scanner sc = new Scanner(System.in);

public List<String> catAndMouse(int x, int y, int z) {
    List<String> results = new ArrayList<String>();
    int q = 0;
    System.out.println("enter the number of queries");
    q = sc.nextInt();
    for (int i = 0; i < q; i++) {
        x = sc.nextInt();
        y = sc.nextInt();
        z = sc.nextInt();
        if (Math.abs(x - z) < Math.abs(y - z)) {
             results.add("Cat A");
        } else if (Math.abs(x - z) > Math.abs(y - z)) {
             results.add("Cat B");
        } else {
             results.add("Mouse C");
        }
    }


    return results;
}

public static void main(String[] args) {
    int x = 0, y = 0, z = 0;
    Main tj = new Main();
    List<String> results =  tj.catAndMouse(x, y, z);
    System.out.println(results.toString());
}

}

输入查询数2 1 2 3 1 3 2 [猫B，鼠标C]

Answer 2

是因为这段代码-

System.out.println("enter the number of queries");
        q = sc.nextInt();
        for (int i = 0; i < q; i++) {
            x = sc.nextInt();
            y = sc.nextInt();
            z = sc.nextInt();
        }

您要向用户询问多个查询，然后在循环中多次询问x，y和z，而不在循环内处理值。

您的catAndMouse方法的唯一责任应该是在给定输入参数的情况下给出结果。因此，将输入留给主类。您可以像下面这样简化代码：

public class TomAndJerry {
    private static Scanner sc = new Scanner(System.in);

    public String catAndMouse(int x, int y, int z) {
        String result;
        if (Math.abs(x - z) < Math.abs(y - z)) {
            result = "Cat A";
        } else if (Math.abs(x - z) > Math.abs(y - z)) {
            result = "Cat B";
        } else {
            result = "Mouse C";
        }
        return result;
    }

    public static void main(String[] args) {

        System.out.println("enter the number of queries");
        int q = sc.nextInt();
        int x , y , z ;
        String result;
        TomAndJerry tj = new TomAndJerry();
        for (int i = 0; i < q; i++) {
            x = sc.nextInt();
            y = sc.nextInt();
            z = sc.nextInt();
            result = tj.catAndMouse(x, y, z);
            System.out.println(result);
        }
    }
}

如何使用扫描仪获得多个输入的多个输出

2 个答案: