计算数学方程字符串中的字母和单词

Question

So, I have written a method named varCount(), shown below, that would count the number of letter variables in a math equation.

// This method counts the number of variables in the equation
    /* Note: If a variable is a word or a substring with length greater than 1,
       then count it as a whole */
    public int varCount(){
        int count = 0;
        for (int i = 0; i < getEquation().length(); i++){
            if ((getEquation().charAt(i) >= 65 && getEquation().charAt(i) <= 90)
            || (getEquation().charAt(i) >= 97 && getEquation().charAt(i) <= 122)){
                count++;
            }
        }
        return count;
    }

This method is made inside a class, so the getter method of the private equation string attribute was used here.

One example of what should result from this was, if you are using the method on a string x + 2 = 3 , there would be 1 letter variable, which was x , that's counted. After testing, the method returned the expected int value and was functional to that extent.

What I really want to accomplish is, if a user ever puts in a variable like num , then the method should still work like the second comment.

On the first attempt, I thought that, since it would be a word like num , then, if the previous character was counted as a letter, then the whole word would be counted by counting just that first letter, like so:

// This method counts the number of variables in the equation
    /* Note: If a variable is a word or a substring with length greater than 1,
       then count it as a whole */
    public int varCount(){
        int count = 0;
        for (int i = 0; i < getEquation().length(); i++){
            if ((getEquation().charAt(i) >= 65 && getEquation().charAt(i) <= 90)
            || (getEquation().charAt(i) >= 97 && getEquation().charAt(i) <= 122)){
                if (i != 0 && (getEquation().charAt(i-1) >= 65 && getEquation().charAt(i-1) <= 90)
                || (getEquation().charAt(i-1) >= 97 && getEquation().charAt(i-1) <= 122)){
                    continue;
                }
                count++;
            }
        }
        return count;
    }

The issue with this was that the method only resulted in an IndexOutOfBoundsException .

The next attempts were either modifications or replacements of the method body with the regex package, which returned 0 using the Matcher class's groupCount() method, like so:

// This method counts the number of variables in the equation
    /* Note: If a variable is a word or a substring with length greater than 1,
       then count it as a whole */
    public int varCount(){
        Pattern alphaRange = Pattern.compile("[a-zA-Z]");
        Matcher match = alphaRange.matcher(getEquation());
        System.out.println(match.groupCount());
        return match.groupCount();
    }

What am I missing or going wrong about this method?

Answer 1

Okay, I had modified the method to use the regex package's Pattern and Matcher class and managed to get the program to function exactly as intended.

// This method counts the number of variables in the equation
    /* Note: If a variable is a word or a substring with length greater than 1,
       then count it as a whole */
    public int varCount(){
        int count = 0;
        Pattern pattern = Pattern.compile("(a-zA-Z)*");
        Matcher match = pattern.matcher(getEquation());
        if (match.find()){
            count = match.groupCount();
        }
        return count;
    }

I had not realized that the groupCount() method needed to be processed if a match was found, logically. I had also used parentheses instead of open brackets for a more specific range.

What I am reminded about, though, is that there could be times when we need another pair of eyes. Thank you!