How to output an arithmetic expression with unary operator?

Question

I have an arithmetic expression(a string) with unary operator, and I want to put each element in an array. For example: -3+4.2*5==>output should be: -3, +, 4.2 ,* , 5(not -,3,+,4.2, *, 5) 3+-5 ==> output should be: 3,+,-5(with the unary operator) (3/(5-8)+18)2==>output should be: (,3,/,(,5,-,8,),+,18,),,2

Here is the code I tried so far, and the output is 3,+,-,5, which didn't put the unary operator in the front of a digit.

My question is how to put each element properly in an array.

     public class Test2 {

/**
 * @param args the command line arguments
 */
public static void main(String[] args) {
    System.out.println("Input:");
    Scanner in = new Scanner(System.in);
    String input = in.nextLine();
    String[]  arr1= splitInfixExpression(input);
    for(int i=0;i<arr1.length;i++)
    {
        System.out.println(arr1[i]);
    }

    }
  priva te static boolean isOperandChar(final char c) {
    return Character.isDigit(c) || c == '.';
   }
  private static boolean isParenthesis(final char c) {
return c=='('||c==')';
}




    private static String[] splitInfixExpression(final String input) {
    final List<String> postfixExpression = new ArrayList<>();
    boolean encounteredOperandStart = false;
    String currentOperand = "";
    for (final char c : input.toCharArray()) {
        if (encounteredOperandStart) {
            if (isOperandChar(c)) {
                currentOperand += c;
            } 


                postfixExpression.add(currentOperand);
                postfixExpression.add(String.valueOf(c));
                currentOperand = "";
                encounteredOperandStart = false;

        } else {
            if (isOperandChar(c)) {
                encounteredOperandStart = true;
                currentOperand += c;
            }

            else if(isParenthesis(c)) {
            postfixExpression.add(String.valueOf(c));
            //currentOperand = "";
           encounteredOperandStart=false;
        }  
            else{
           postfixExpression.add(String.valueOf(c));
            //currentOperand = "";
           encounteredOperandStart=false;
       }                

        }
    }
    if (!currentOperand.isEmpty()) {
        postfixExpression.add(currentOperand);
    }
    return postfixExpression.toArray(new String[postfixExpression.size()]);
}}

This is a pretty intense problem for logic. I wouldn't even want to try it, to be honest. I would use regular expressions. — aliteralmind, Mar 30 '14 at 23:02
Also see http://stackoverflow.com/questions/3422673/evaluating-a-math-expression-given-in-string-form — aliteralmind, Mar 30 '14 at 23:03
aliteralmind, this is not a me thematic evaluation question, it is how to put a string into an array. — Sophie, Mar 30 '14 at 23:44
Of course. The first step in these linked questions is obviously getting the formula-string pieces into an array. Solving is the easy(er) part. — aliteralmind, Mar 30 '14 at 23:58
@Sophie You may not realize it, but the question is about parsing arithmetic expressions. — user207421, Mar 31 '14 at 02:15
@DavidWallace Because you can't get the unary operator into the right place without doing a proper parse, and for that you need a correct arithmetic expression parsing technique, such as recursive descent, or the Dijkstra shunting-yard algorithm. These algorithms have been known for many decades. *Ad hoc* techniques do not work. — user207421, Mar 31 '14 at 02:19
Whether or not that's true, that doesn't mean that "how do I break this string into an array" is automatically a duplicate of "how do I calculate a numeric value for this expression". These two questions have a lot of common ground, but they're not duplicates of each other. Voting to reopen. — Dawood ibn Kareem, Mar 31 '14 at 02:22

Stephen Buttolph · Answer 1 · 2014-03-31T03:03:29.760

0

I think this is want you want? :)

public static List<String> splitInfixExpression(String someString){
    List<String> someList = new ArrayList<String>();
    String tempString = "";
    for (int i = 0; i < someString.length(); i++){
        if (Character.isDigit(someString.charAt(i)) || (someString.charAt(i) == '-' && someString.length() == 0) || someString.charAt(i) == '.'){
            tempString += String.valueOf(someString.charAt(i));
            tempString = tempString.trim();
        }
        else{
            if (tempString.length() > 0){
                someList.add(tempString);
            }
            tempString = String.valueOf(someString.charAt(i));
            if (someList.size() > 0 && Character.isDigit(someList.get(someList.size() - 1).charAt(someList.get(someList.size() - 1).length() - 1))){
                someList.add(tempString);
                tempString = "";
            }
            else if (tempString.trim().length() > 0 && ((!tempString.equals("-"))) && ((!tempString.equals("+")))){
                someList.add(tempString);
                tempString = "";
            }
        }
    }
    someList.add(tempString);

    return someList;
}

edited Mar 31 '14 at 03:03

answered Mar 30 '14 at 23:05

Stephen Buttolph

643
8
16

Can you show some example inputs and outputs please? I'd be interested in seeing it with `(3/(5--8)+18)2` (five minus negative 8). – aliteralmind Mar 30 '14 at 23:07
It doesn't. No parentheses, no operator precedence. – user207421 Mar 30 '14 at 23:15
this only works for -3+4, 3+-4, but for 3-+4, it is not working, thank you StephenB – Sophie Mar 30 '14 at 23:22
1

@Sophie I imagine it could be easily adapted to include those cases. Have a look at the lines where characters or strings are compared to `-` and maybe extend them to include `+`. – Dawood ibn Kareem Mar 31 '14 at 00:17
@Sophie for 3-+4 do you want to output to be 3,-,4? Or do you want it to be 3,-,+,4? Or something else? – Stephen Buttolph Mar 31 '14 at 00:52
The output of 3-+4 should be 3,-,+4 and the output of 3+-4 should be 3,+,-4. Thank you – Sophie Mar 31 '14 at 01:43
@DavidWallace It could only be 'adapted' by completely rewriting it. A stack is necessary for a start, either explicitly or via recursion. – user207421 Mar 31 '14 at 02:08
@StephenB did you test this? I think there's a mistake near the top. – Dawood ibn Kareem Mar 31 '14 at 02:10
@DavidWallace Using a for-each loop I iterate through the returned array and it works fine for me. – Stephen Buttolph Mar 31 '14 at 02:15
Then I stand corrected. Maybe I didn't see a mistake after all. But it always inspires a lot of confidence to see "This might work?" at the top of an answer. – Dawood ibn Kareem Mar 31 '14 at 02:19
Haha well, that was more at the fact that i've misunderstood what Sophie asked at first, not that the code didn't do what i wanted. – Stephen Buttolph Mar 31 '14 at 02:22

How to output an arithmetic expression with unary operator?

1 Answers1