如何重新创建下表？

Question

大家好，我正在尝试为作业重新创建上面的表格，但由于有关 && 函数的错误而卡住了。请注意，我们不允许使用数组，并且仅限于“switch”和“If”这是我到目前为止所得到的：if ((BEAK_MM == 1) && (CLAW_MM == 0) && (COLOR = "Grey")) {   System.out.println ("The type of bird is A.");}else if ((BEAK_MM == 2) && (CLAW_MM == 1) && (COLOR = "Grey"))    System.out.println ("The type of bird is A.");}else if ((BEAK_MM == 3) && (CLAW_MM == 2) && (COLOR = "Grey"))    System.out.println ("The type of bird is A.");}else if ((BEAK_MM == 4) && (CLAW_MM == 3) && (COLOR = "Grey"))    System.out.println ("The type of bird is A.");}else if ((BEAK_MM <= 4.5) && (CLAW_MM == 4) && (COLOR = "Grey"))    System.out.println ("The type of bird is A.");}

Answer 1

假设BEAKandCLAW是整数并且COLOR是 a String，那么表格可以更像：

// must be grey

if ("Grey".equals(COLOR)) {

  if ((BEAK_MM == 1 && CLAW_MM == 0)

      || (BEAK_MM == 2 && CLAW_MM == 1)

      || (BEAK_MM == 3 && CLAW_MM AW == 2)

      || (BEAK_MM == 4 && CLAW_MM == 3)

      || ( (BEAK_MM == 4 || BEAK_MM == 5) && CLAW_MM == 4))) {

     System.out.println("The bird is the word");

  }

}

这里的逻辑是，根据表格，鸟A型一定是灰色的。然后有一些关于喙和爪类型的特定检查，但如果它不是灰色，则它不是 A 型鸟。也不需要所有 OP 堆叠的“else if”语句。

有许多其他方法可以解决问题空间，但我没有数组的限制，因此大概没有其他有用的数据结构，例如Listor Set。

正如@Nicholas K 所指出的，String对象必须与.equals().

我也会把这些东西移到一个方法上 isBirdTypeA(int beak, int claw, String color) { ... }

@Test

public void testBirds()

{

    final String G = "Grey";

    final String P = "Pink";

    assertTrue(isBirdTypeA(1, 0, G));

    assertFalse(isBirdTypeA(1, 0, P));

    assertTrue(isBirdTypeA(2, 1, G));

    assertTrue(isBirdTypeA(3, 2, G));

    assertTrue(isBirdTypeA(4, 3, G));

    assertTrue(isBirdTypeA(4, 4, G));

    assertTrue(isBirdTypeA(5, 4, G));

    assertFalse(isBirdTypeA(4, 5, G));

    assertFalse(isBirdTypeA(4, 0, G));

    assertFalse(isBirdTypeA(1, 1, G));

}

private static boolean isBirdTypeA(int beak, int claw, String color)

{

    if ("Grey".equals(color)) {

        if ((beak == 1 && claw == 0)

            || (beak == 2 && claw == 1)

            || (beak == 3 && claw == 2)

            || (beak == 4 && claw == 3)

            || ( (beak == 4 || beak == 5) && claw == 4)) {

                return true;

            }

    }

    return false; 

}

public static void main(String[] args) {

  int BEAK_MM = Integer.parseInt(args[0]);

  int CLAW_MM = Integer.parseInt(args[1]);

  String COLOR = args[2];

  if (isBirdTypeA(BEAK_MM, CLAW_MM, COLOR)) {

    System.out.println("The type of bird is A");

  }

}

$ java BirdBeak 1 0 Gray

鸟的类型是 A