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I likeThatMeow
Asked: 2020-09-06 16:28:31 +0800 CST 2020-09-06 16:28:31 +0800 CST

Error in a code when data is generalized

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I have a question about the following code, the idea is that if given as input:

a + b + c + d + ... + z

Gives as output:

correct expression

If given as input:

a + b + c ++ b + b b...z

Gives as output:

wrong expression

The program works fine if it is given:

a+b

But there is an error when I enter it

a+b+c+d

Well, it gives me as output:

incorrect expression

When it is correct.

public class Ejercicio5sep {

public static void main(String[] args) {

    Scanner x = new Scanner(System.in);
    LinkedList pila = new LinkedList();
    int correcto = 0;

    System.out.println("Cuantos elementos quiere introducirle a la pila?");
    int n = x.nextInt();

    for (int i = 0; i < n; i++) {
        System.out.println("introduzca su elemento");
        String e = x.next();
        pila.push(e);
    }
    if (pila.size() % 2 != 0) {

        float d = (float) pila.size() / 2f + 0.5f;

        for (int i = 0; i < pila.size() / 2; i++) {
            String a = (String) pila.pop();
            String b = (String) pila.pop();

            if (false == a.equals("+") && b.equals("+")) {
                correcto++;
            }
        }

        String c = (String) pila.pop();

        if (false == c.equals("+"))//(*)
        {
            correcto++;
        }
        //si input: a+b+c+d
        System.out.println(correcto);//dice que correcto==2, entonces 
        //no hizo la instruccion (*), ?porque

        if (correcto == d) {
            System.out.println("La expresion es correta");
        } else {
            System.out.println("La expresion no es correta");
        }
    } else {
        System.out.println("expresion incorrecta :(");
    }
java

1 Answers

  1. Best Answer

    Good morning, I propose an alternative that I think is better understood.

    Concept

    Notice the pattern of your correct solutions:

    a+b

    a+b+c+d

    a+b+c+...+z

    All these solutions have in common that the odd elements are an operation (+). Since you keep everything in a stack, we could check if the expression is correct or not, analyzing that the even elements are different from "+" and the odd ones are equal to "+".

    Code

    public class Ejercicio5sep {

public static void main(String[] args) {

    Scanner x = new Scanner(System.in);
    LinkedList pila = new LinkedList();
    boolean correcto = true; //suponemos que es correcto desde el principio

    System.out.println("Cuantos elementos quiere introducirle a la pila?");
    int n = x.nextInt();

    for (int i = 0; i < n; i++) {
        System.out.println("introduzca su elemento");
        String e = x.next();
        pila.push(e);
    }
    //a partir de aquí cambiamos el código
    if(pila.size() % 2 != 0)  //Si el tamaño de la pila es par, la expresión es incorrecta
    {
       for(int i = 0; i < pila.size(); i++)
       {
          String a = pila.get(i).toString();
          //si es elemento par y no es "+"
          if(i % 2 == 0 && !a.equals("+"))         
          {
             continue; //continuamos con la siguiente iteración
          }
          else
          {
             if(i % 2 != 0 && a.equals("+")) //es impar e igual a "+"
             {
                 continue;
             }
             else
             {
                 correcto=false;
                 break; //salimos del for, no es necesario seguir
             }
          }
       }
    }
    else
    {
       correcto = false;
    }
   if(correcto)
   {
      System.out.println("La expresion es correta");
   }
   else
   {
      System.out.println("La expresion no es correta");
   }
  }
}

    I use stack.get(i) instead of stack.pop() because I understand that stack.pop() will modify the size of the stack and therefore the value returned by stack.size() .

    This way of setting up the algorithm saves you from having to analyze the entire expression if it is incorrect, since as soon as it finds something incorrect, it stops searching.

    I hope it helps you and is easy to understand.

    • 1