Using unary representation with addition java string

Question

I need to write a code that takes a string input and turns it, or something to that effect, into a valid unary equation with addition to verify if it is valid. I'm confused, could anyone point me in the direction of understanding this? An example would be: 111+1111=11111+1+1 is the statement 3+4=5+1+1 which is valid. My other question would be how to use stacks to do unary operations.

Answer 1

If you are limited to this language then your can write a simple parser using a number of methods. You can first split your String

String[] parts = eqn.split("=");

Then split each part:

String[] left = parts[0].split("\\+");
String[] right = parts[1].split("\\+");

Now you can simply count the lengths of the strings:

int leftside = 0;
for (String l : left) {
    leftside += l.length;
}

and so on for the right side and see if they are equal.

There are many other ways you can tackle this.

Basically you have to write a parser. You might want to use a character-by-character scanner, or use regexes or a generic tool such as ANTLR. It depends on your final goal and whether it is likely to change. Are you likely to have characters other than 1, +, =, for example?

My guess is this is homework. And I guess that you are expected to read character-by-character and push() each character on a stack. Then you have to pop() the stack when you encounter certain conditions.I'm not going to do this for you...

Answer 2

Another possible solution.

String input = "111+1111=11111+1+1";
    String[] parts = input.split("=");

    Pattern pattern = Pattern.compile("1");
    Matcher  matcherLeft = pattern.matcher(parts[0]);
    Matcher matcherRight = pattern.matcher(parts[1]);
    int leftTotal = 0;
    while (matcherLeft.find())
        leftTotal++;

    int rightTotal = 0;
    while (matcherRight.find())
        rightTotal++;

    if(leftTotal == rightTotal)
        System.out.println("Valid");
    else
        System.out.println("Invalid");

Starts out by splitting the string in to the left and right side of the equations. Then simply counts the number of 1's in each part and does a comparison. There's definitely better ways to do this, but with this example it's pretty easy to see what is going on.