Question

I am creating a program and part of the program receives an expression with the following conditions:

• Allowed chars: “1234567890+-*/() ” — notice the space at the end of them

• Must be balanced

• Starts and ends with “(” and “)”

• The separation is a single white space between each operator, operand or/and parenthesis:

I have done the first two conditions, but I am stuck with the third one.

Examples:

( 1 + ( ( 22 – 322 ) * 4 / 5 ) ) <<< Valid

( 1+ ( ( 22 – 322 ) * 4 / 5 ) ) <<< Invalid

( 1 + (( 22 – 322 ) * 4 / 5 ) ) <<< Invalid

( 1 + ( ( 22 – 322 )* 4 / 5 ) ) <<< Invalid

Here is what I have done so far:

public static boolean isSSWS(String e) {

    boolean valid = false;
    int indexOfWS = 0;
    Character preChar = '\0';
    Character postChar = '\0';

    while(!valid && (indexOfWS < e.length())) {
        indexOfWS = e.indexOf(" ", indexOfWS + 1);
        preChar = e.charAt(indexOfWS - 1);
        postChar = e.charAt(indexOfWS + 1);

        if(preChar == ' ' || postChar == ' ') {
            return false;
        }
    }

    return true;
}

isSSWS, is *S*eparated with *S*ingle *W*ite *S*pace

indexOfWS, index of white space

Answer 1

Edited, I didn’t notice that numbers shouldn’t have spaces. Bit messy, but works for all inputs I’ve tried.

public static boolean isSSWS(String s) {
    boolean isWhiteSpace = false;
    boolean isPreviousCharDigit = false;

    for (char c : s.toCharArray()) {
        if (!isPreviousCharDigit || !Character.isDigit(c)) {
            if (c == ' ' && !isWhiteSpace) {
                return false;
            }
            if (c != ' ' && isWhiteSpace) {
                return false;
            }
            isWhiteSpace = !isWhiteSpace;
        }
        isPreviousCharDigit = Character.isDigit(c);
    }
    return true;
}