RPN Rechner mit Stacks

Question

Dies ist ein Code, den ich von meinem Lehrbuch bekommen habe, das ein Rechner ist, der Ausdrücke löst, die als Postfix-Notation eingegeben werden. Ich habe mich gefragt, ob es eine Möglichkeit gibt, den Stack weiter zu verwenden, anstatt den Stack neu zu starten, sobald das Programm einen Ausdruck ausgewertet hat.

import java.util.Scanner; 
    import java.util.Stack; 

    public class RPN2 
    { 
    private Stack<Integer> stack; 

    public RPN2() 
     { 
      stack = new Stack<Integer>(); //creates stack 
     } 

    public static void main(String[] args) 
     { 
      String expression, again; 
      int result; 

      Scanner keyboard = new Scanner(System.in); 

      do 
      { 
       RPN2 evaluator = new RPN2(); 
       System.out.println("Enter a valid post-fix expression one token " + 
            "at a time with a space between each token (e.g. 5 4 + 3 2 1 - + *)"); 
       System.out.println("Each token must be an integer or an operator (+,-,*,/)"); 
       expression = keyboard.nextLine(); 

       result = evaluator.evaluate(expression); 
       System.out.println(); 
       System.out.println("That expression equals " + result); 

       System.out.print("Evaluate another expression [Y/N]? "); 
       again = keyboard.nextLine(); 
       System.out.println(); 
      } 
      while (again.equalsIgnoreCase("y")); 
     } 

    public int evaluate(String expr) 
     { 
      int op1, op2, result = 0; 
      String token; 
      Scanner parser = new Scanner(expr);  

      while (parser.hasNext())   
      { 
       token = parser.next();   

       if (isOperator(token)) //if operator pop 
       { 
        op2 = (stack.pop()).intValue(); 
        op1 = (stack.pop()).intValue(); 
        result = evaluateSingleOperator(token.charAt(0), op1, op2);  // 
        stack.push(new Integer(result)); 
       } 
       else 
        stack.push(new Integer(Integer.parseInt(token)));  
      } 

      return result; 
     } 

    private boolean isOperator(String token) 
     { 
      return (token.equals("+") || token.equals("-") || 
        token.equals("*") || token.equals("/") || token.equals("%")); 

     } 

    private int evaluateSingleOperator(char operation, int op1, int op2) 
     { 
      int result = 0; 

      switch (operation) 
      { 
       case '+': 
        result = op1 + op2; 
        break; 
       case '-': 
        result = op1 - op2; 
        break; 
       case '*': 
        result = op1 * op2; 
        break; 
       case '/': 
        result = op1/op2; 
        break; 
       case '%': 
        result = op1 % op2; 
        break; 
      } 

      return result; 
     } 

} 

Answer 1

Ja, Sie können, siehe Kommentare:

public RPN2() { 
    stack = new Stack<>(); //creates stack 
} 

void clearStack(){ //add clear stach method 
    stack.clear(); 
} 

public static void main(String[] args) 
{ 
    String expression, again; 
    int result; 

    Scanner keyboard = new Scanner(System.in); 
    RPN2 evaluator = new RPN2(); //move out of the do loop 

    do 
    { 
     evaluator.clearStack();//use clear stack method 
     //rest of the code omitted (no change in it)