java数学运算符

Question

我有这样一种方法，它将生成一个随机数学表达式，求解它并输出变量的答案：

public int Nov2()
{
    char[] ops = new char[] {'+', '-', '*', '/'};
    int i = rand.nextInt(4-0) + 0;
    char op1 = ops[i];

    int novnum1 = rand.nextInt(101-1) + 1;

    int novnum2 = rand.nextInt(101-1) + 1;

    int nov2result = 0;

    switch(op1) {
        case '+': nov2result = novnum1 + novnum2; break;
        case '-': nov2result = novnum1 - novnum2; break;
        case '*': nov2result = novnum1 * novnum2; break;
        case '/': nov2result = novnum1 / novnum2; break;
    }

    String nov2Exp = novnum1 + " " + op1 + " " + novnum2 + " = ";

    Nov2resstor = nov2result;

    setContentView(R.layout.gameview);

    TextView display = (TextView) findViewById(R.id.exp);

    display.setText(nov2Exp);

    return nov2result;
}

在我的下一个方法中，我将如何对包含两个以上术语的表达式使用相同的方法，而不必编写非常复杂的if语句：

public int Eas3()
{
    char[] ops = new char[] {'+', '-', '*', '/'};
    int i = rand.nextInt(4-0) + 0;
    char op1 = ops[i];
    i = rand.nextInt(4-0) + 0;
    char op2 = ops[i];

    int easnum1 = rand.nextInt(101-1) + 1;

    int easnum2 = rand.nextInt(101-1) + 1;

    int easnum3 = rand.nextInt(101-1) + 1;

    int eas3result = 0;


    if (op1 == '+' && op2 == '+')
    {
        eas3result = ((easnum1 + easnum2) + easnum3);
    }
    else if (op1 == '+' && op2 == '-')
    {
        eas3result = ((easnum1 + easnum2) - easnum3);
    }
    else if (op1 == '+' && op2 == '*')
    {
        eas3result = ((easnum1 + easnum2) * easnum3);
    }
    else if (op1 == '+' && op2 == '-')
    {
        eas3result = ((easnum1 + easnum2) - easnum3);
    } 
.../

我有一些方法可以对2,3,4,5和6执行此操作，所以我的if语句将使用此方法变得非常大

有什么想法吗

Answer 1

是的，另一种方法是编写命令对象：

public interface Command<V> {
    V execute(Object ... args); 
}

您将编写一个实现此接口的对象：

public class AdditionCommand implements Command<Double> {
    public Double execute(Object ... args) {
        Double x = (Double)args[0];
        Double y = (Double)args[1];
        return x+y;    
    }
}

现在，您可以使用运算符在地图中查找：

Map<String, Command> opsLookup = new HashMap<String, Command>() {{
   opsLookup.put("+", new AddCommand<Number>());
   opsLookup.put("-", new SubtractCommand<Number>());
}};

不需要开关

Answer 2

你要找的是composite pattern。定义一个抽象的Expression基类并派生它

这些类必须实现返回结果的evaluate()方法

一个子类将是返回其值的常量，另一个子类将是一个二进制表达式，如加号、减号等。evaluate()方法将添加/减去/etc求值子表达式的结果

然后，您可以从其他表达式中构建任意表达式，然后在不使用if条件的情况下对其求值

Answer 3

Check out this MathEval class I found online它将为您计算表示方程式的字符串

mySolver = new MathEval();
double answer = mySolver.evaluate(equation);

Answer 4

您可以使用内置的Javascript引擎

import javax.script.ScriptEngineManager;
import javax.script.ScriptEngine;

public class Test 
{
  public static void main(String[] args) throws Exception
  {
       ScriptEngineManager mgr = new ScriptEngineManager();
        ScriptEngine engine = mgr.getEngineByName("JavaScript");
        String foo = "40+2";
        System.out.println(engine.eval(foo));
    } 
}

Answer 5

使用递归如何：

int num(int numberOfOperands, int current){
    if(numberOfOperands<=0) return current;
    switch(rand.nextInt(4)){
        case 0: return num(numberOfOperands-1, current + (rand.nextInt(100)+1)); break;
        case 1: return num(numberOfOperands-1, current - (rand.nextInt(100)+1)); break;
        case 2: return num(numberOfOperands-1, current * (rand.nextInt(100)+1)); break;
        case 3: return num(numberOfOperands-1, current / (rand.nextInt(100)+1)); break;
    }

}

int num(int numberOfOperands) throws Exception{
    if(numberOfOperands <=0) 
        throw new Exception("invalid number of operands: "+numberOfOperands);
    return num(numberOfOperands, rand.nextInt(100)+1);
}

当然，这会忽略操作的优先级

Answer 6

您可以使用正在使用的变量创建一个字符串，如下所示：

String temp = "(" + easnum1 + op1 + easnum2 + ")" + op2 + easnum3;

之后，您可以使用ScriptEngineManager类将javascript用作引擎，以便使用eval方法

ScriptEngineManager manager = new ScriptEngineManager();
ScriptEngine engine = manager.getEngineByName("js");        
Object result = engine.eval(temp);

此方法执行计算并返回结果

希望这有帮助