infix to post fix convertion

Here is an example of converting infix expression to its postfix. The code accepts the expression from the user and display its postfix expression.

import java.util.Scanner;
import java.util.Stack;

public class InfixToPostfix {
  private String infix;
  private String postfix = "";

  public void convertString(String a){
    String str = "";
    infix = a;
    Stack<String> stack = new Stack<String>();

    for(int i = 0; i < infix.length(); i++){
      str = infix.substring(i,i+1);
      if(str.matches("[a-zA-Z]|\\d"))
        postfix += str;
      else if (isOperator(str)){
        if (stack.isEmpty()){
          stack.push(str);
        }
        else{
          String stackTop = stack.peek();
          while (getPrecedence(stackTop,str).equals(stackTop)&& !(stack.isEmpty())){
            postfix += stack.pop();
            if (!(stack.isEmpty()))
              stackTop = stack.peek();
          }
          stack.push(str);
        }
      }
    }
    while(!(stack.isEmpty()))
      postfix += stack.pop();
    System.out.println("Postfix of the expression is: " +  postfix);
  }

  private boolean isOperator(String ch){
    String operators = "*/%+-";
    if (operators.indexOf(ch) != -1)
      return true;
    else
      return false;
  }

  private String getPrecedence(String op1, String op2){
    String multiplicativeOps = "*/%";
    String additiveOps = "+-";
    if ((multiplicativeOps.indexOf(op1) != -1) && (additiveOps.indexOf(op2) != -1))
      return op1;
    else if ((multiplicativeOps.indexOf(op2) != -1) && (additiveOps.indexOf(op1) !=  -1))
      return op2;
    else if((multiplicativeOps.indexOf(op1) != -1) && (multiplicativeOps.indexOf(op2) != -1))
      return op1;
    else
      return op1;
  }
  public static void main(String[] args) {

    System.out.println("Enter an expression in the Infix form:");
    Scanner scanner = new Scanner(System.in);

    String expression = scanner.nextLine();
    new InfixToPostfix().convertString(expression);


  } 
}