Java 大型开关箱（60种不同的箱）-如何减少？或者改进代码
所以我有一个开关盒，它以厘米为单位输入一条边的长度，然后选择机器人轮子旋转的时间，以毫秒为单位 用户输入25选择案例25并将时间设置为2000ms 但是有60个不同的可能案例两次，我很困惑如何缩小这个范围？ 如果可能的话Java 大型开关箱（60种不同的箱）-如何减少？或者改进代码,java,eclipse,ide,switch-statement,Java,Eclipse,Ide,Switch Statement,所以我有一个开关盒，它以厘米为单位输入一条边的长度，然后选择机器人轮子旋转的时间，以毫秒为单位 用户输入25选择案例25并将时间设置为2000ms 但是有60个不同的可能案例两次，我很困惑如何缩小这个范围？ 如果可能的话 import java.util.ArrayList; import java.util.Scanner; import edu.cmu.ri.createlab.terk.robot.finch.Finch; public class assignment2 { s
import java.util.ArrayList;
import java.util.Scanner;
import edu.cmu.ri.createlab.terk.robot.finch.Finch;
public class assignment2 {
static String shape,largestShapeName;
static int side1,side2,side3;
static int triangleCount=0,rectangleCount = 0;
static int area,largestShape = 0;
static int largestSide1,largestSide2,largestSide3;
static int angle1,angle2,angle3;
static boolean possible=false;
static Scanner shapescanner = new Scanner(System. in);
static Scanner sidescanner = new Scanner(System. in);
static Finch myfinch = new Finch();
static ArrayList<String> arrayShape = new ArrayList<String>();
public static void main(String args[])
{
System.out.println("Please Enter The shape you want to Draw");
shape = shapescanner.nextLine().toUpperCase();
while (!"Q".equals(shape)){
while (!"Q".equals(shape) && !"R".equals(shape) && !"T".equals(shape) ) {
System.out.println(shape +" is an incorrect Option Please Enter again");
shape = shapescanner.nextLine().toUpperCase();
}
if ("R".equals(shape)) {
Rectangle();
} else if ("T".equals(shape)) {
Triangle();
}
if (area > largestShape) {
if ("R".equals(shape)) {
largestSide1=side1;
largestSide2=side2;
largestShapeName= "R";
}else {
largestSide1=side1;
largestSide2=side2;
largestSide3=side3;
largestShapeName= "T";
}
System.out.println("Please Enter The shape you want to Draw");
shape = shapescanner. nextLine().toUpperCase();
}else {
System.out.println("Please Enter The shape you want to Draw");
shape = shapescanner. nextLine().toUpperCase();
}
}
if ("Q".equals(shape)) {
for (int i = 0; i < arrayShape.size(); i++) {
System.out.println(arrayShape.get(i));
}
myfinch.quit();
System.exit(0);
}
}
public static void Rectangle () {
int side1time = 0;
int side2time = 0;
System.out.println("Please Enter The Length of first side");
side1 = Integer.parseInt(sidescanner. nextLine());
while (side1>80 || side1<20) {
System.out.println(side1 +" is an incorrect Size Please Enter again");
side1 = Integer.parseInt(sidescanner. nextLine());
}
System.out.println("Please Enter The Length of second side");
side2 = Integer.parseInt(sidescanner. nextLine());
while (side2>80 || side2<20) {
System.out.println(side2 +" is an incorrect Size Please Enter again");
side2 = Integer.parseInt(sidescanner. nextLine());
}
switch (side1) {
case 20: side2time = 2000 ;
break;
case 21: side2time = 2000 ;
break;
case 22: side1time = 2100 ;
break;
case 23: side1time = 2200 ;
break;
case 24: side1time = 2300 ;
break;
case 25: side1time = 2400 ;
break;
case 26: side1time = 2500 ;
break;
case 27: side1time = 2600 ;
break;
case 28: side1time = 2700 ;
break;
case 29: side1time = 2800 ;
break;
case 30: side1time = 2900 ;
break;
case 31: side1time = 3000 ;
break;
case 32: side1time = 3100 ;
break;
case 33: side1time = 3200 ;
break;
case 34: side1time = 3300 ;
break;
case 35: side1time = 3400 ;
break;
case 36: side1time = 3500 ;
break;
case 37: side1time = 3600 ;
break;
case 38: side1time = 3700 ;
break;
case 39: side1time = 3800 ;
break;
case 40: side1time = 3900 ;
break;
case 41: side1time = 4000 ;
break;
case 42: side1time = 4100 ;
break;
case 43: side1time = 4200 ;
break;
case 44: side1time = 4300 ;
break;
case 45: side1time = 4400 ;
break;
case 46: side1time = 4500 ;
break;
case 47: side1time = 4600 ;
break;
case 48: side1time = 4700 ;
break;
case 49: side1time = 4800 ;
break;
case 50: side1time = 4900 ;
break;
case 51: side1time = 5000 ;
break;
case 52: side1time = 5100 ;
break;
case 53: side1time = 5200 ;
break;
case 54: side1time = 5300 ;
break;
case 55: side1time = 5400 ;
break;
case 56: side1time = 5500 ;
break;
case 57: side1time = 5600 ;
break;
case 58: side1time = 5700 ;
break;
case 59: side1time = 5800 ;
break;
case 60: side1time = 5900 ;
break;
case 61: side1time = 6000 ;
break;
case 62: side1time = 6100 ;
break;
case 63: side1time = 6200 ;
break;
case 64: side1time = 6300 ;
break;
case 65: side1time = 6400 ;
break;
case 66: side1time = 6500 ;
break;
case 67: side1time = 6600 ;
break;
case 68: side1time = 6700 ;
break;
case 69: side1time = 6800 ;
break;
case 70: side1time = 6900 ;
break;
case 71: side1time = 7000 ;
break;
case 72: side1time = 7100 ;
break;
case 73: side1time = 7200 ;
break;
case 74: side1time = 7300 ;
break;
case 75: side1time = 7400 ;
break;
case 76: side1time = 7500 ;
break;
case 77: side1time = 7600 ;
break;
case 78: side1time = 7700 ;
break;
case 79: side1time = 7800 ;
break;
case 80: side1time = 7900 ;
break;
}
switch (side2) {
case 20: side2time = 2000 ;
break;
case 21: side2time = 2000 ;
break;
case 22: side2time = 2100 ;
break;
case 23: side2time = 2200 ;
break;
case 24: side2time = 2300 ;
break;
case 25: side2time = 2400 ;
break;
case 26: side2time = 2500 ;
break;
case 27: side2time = 2600 ;
break;
case 28: side2time = 2700 ;
break;
case 29: side2time = 2800 ;
break;
case 30: side2time = 2900 ;
break;
case 31: side2time = 3000 ;
break;
case 32: side2time = 3100 ;
break;
case 33: side2time = 3200 ;
break;
case 34: side2time = 3300 ;
break;
case 35: side2time = 3400 ;
break;
case 36: side2time = 3500 ;
break;
case 37: side2time = 3600 ;
break;
case 38: side2time = 3700 ;
break;
case 39: side2time = 3800 ;
break;
case 40: side2time = 3900 ;
break;
case 41: side2time = 4000 ;
break;
case 42: side2time = 4100 ;
break;
case 43: side2time = 4200 ;
break;
case 44: side2time = 4300 ;
break;
case 45: side2time = 4400 ;
break;
case 46: side2time = 4500 ;
break;
case 47: side2time = 4600 ;
break;
case 48: side2time = 4700 ;
break;
case 49: side2time = 4800 ;
break;
case 50: side2time = 4900 ;
break;
case 51: side2time = 5000 ;
break;
case 52: side2time = 5100 ;
break;
case 53: side2time = 5200 ;
break;
case 54: side2time = 5300 ;
break;
case 55: side2time = 5400 ;
break;
case 56: side2time = 5500 ;
break;
case 57: side2time = 5600 ;
break;
case 58: side2time = 5700 ;
break;
case 59: side2time = 5800 ;
break;
case 60: side2time = 5900 ;
break;
case 61: side2time = 6000 ;
break;
case 62: side2time = 6100 ;
break;
case 63: side2time = 6200 ;
break;
case 64: side2time = 6300 ;
break;
case 65: side2time = 6400 ;
break;
case 66: side2time = 6500 ;
break;
case 67: side2time = 6600 ;
break;
case 68: side2time = 6700 ;
break;
case 69: side2time = 6800 ;
break;
case 70: side2time = 6900 ;
break;
case 71: side2time = 7000 ;
break;
case 72: side2time = 7100 ;
break;
case 73: side2time = 7200 ;
break;
case 74: side2time = 7300 ;
break;
case 75: side2time = 7400 ;
break;
case 76: side2time = 7500 ;
break;
case 77: side2time = 7600 ;
break;
case 78: side2time = 7700 ;
break;
case 79: side2time = 7800 ;
break;
case 80: side2time = 7900 ;
break;
}
myfinch.setLED(0, 255, 0, 2000);
myfinch.setWheelVelocities(100,100,side1time);
myfinch.setLED(255, 0, 0, 200);
myfinch.setWheelVelocities(100,0,2000);
myfinch.setWheelVelocities(100,100,side2time);
myfinch.setLED(255, 0, 0, 200);
myfinch.setWheelVelocities(100,0,2000);
myfinch.setWheelVelocities(100,100,side1time);
myfinch.setLED(255, 0, 0, 200);
myfinch.setWheelVelocities(100,0,2000);
myfinch.setWheelVelocities(100,100,side2time);
myfinch.setLED(255, 0, 0, 200);
myfinch.setLED(0, 0, 0, 200);
myfinch.setLED(255, 0, 0, 200);
rectangleCount += 1;
arrayShape.add(shape + ":" + side1 + "," + side2);
System.out.println(arrayShape.get(0));
area = side1 * side2;
}
public static void Triangle() {
}
}
if (side1 == 20) side2time = 2000;
if (side1 > 20 && side1 <= 80) side2time = (side1 - 1) * 100
if (side1 == 20) side2time = 2000;
if (side1 > 20 && side1 <= 80) side2time = (side1 - 1) * 100
这同样适用于另一个switch语句。如果要分配的值可以在以后某个时间点更改，则可以使用键/值对填充映射并使用该映射，而可以将这些值放入文件并读入，因此如果这些值更改，则无需重新编译类：
private final static Map<Integer, Integer> SIDE2TIMEMap = new HashMap<>();
static {
SIDE2TIMEMap.put(Integer.valueOf(20), Integer.valueOf(2000));
SIDE2TIMEMap.put(Integer.valueOf(21), Integer.valueOf(2000));
// [...]
SIDE2TIMEMap.put(Integer.valueOf(80), Integer.valueOf(7900));
}
如果要分配的值可以在以后某个时间点更改，则可以使用键/值对填充映射并使用该映射。您可以将这些值放入一个文件并读入，因此，如果这些值更改，则无需重新编译类：
private final static Map<Integer, Integer> SIDE2TIMEMap = new HashMap<>();
static {
SIDE2TIMEMap.put(Integer.valueOf(20), Integer.valueOf(2000));
SIDE2TIMEMap.put(Integer.valueOf(21), Integer.valueOf(2000));
// [...]
SIDE2TIMEMap.put(Integer.valueOf(80), Integer.valueOf(7900));
}
非常感谢。这确实适用于给定的值，但是，我必须承认它们是占位符，因为必须测试每个距离，并通过更改ms值使其达到正确的距离来测试20cm、21cm、22xm+。实际值并不像side1-1*100那样线性，这将与这些值一起工作values@AliB2512请在你的帖子中包括这些要求和所有其他要求。重写代码取决于您想要/需要的语义。如果你不解释语义，我们就无法在回答中解释它们。@AliB2512要么有一个计算边时间的公式，要么每个值都是任意的。在第一种情况下，只需在代码中包含公式，在第二种情况下，您必须手动写入所有可能的值-使用开关可以被映射替换，但仍有60行代码。@assylias，不幸的是，它们都是通过实验得出的任意值，所以我想我也无能为力吧，我有那么多行代码谢谢！这确实适用于给定的值，但是，我必须承认它们是占位符，因为必须测试每个距离，并通过更改ms值使其达到正确的距离来测试20cm、21cm、22xm+。实际值并不像side1-1*100那样线性，这将与这些值一起工作values@AliB2512请在你的帖子中包括这些要求和所有其他要求。重写代码取决于您想要/需要的语义。如果你不解释语义，我们就无法在回答中解释它们。@AliB2512要么有一个计算边时间的公式，要么每个值都是任意的。在第一种情况下，只需在代码中包含公式，在第二种情况下，您必须手动写入所有可能的值-使用开关可以被映射替换，但仍有60行代码。@assylias，不幸的是，它们都是通过实验得出的任意值，所以我想我无能为力吧，有很多行代码边时间永远不会改变，它们几乎就像一个查找表，如果你想移动20厘米，轮子应该移动x数量边时间永远不会改变，它们几乎就像一个查找表，如果你想移动20厘米，轮子应该移动x数量