Question
In this assignment, do not use any auxiliary data structures (in particular, do not use a stack)
Write a recursive method, inx, to transform a postx expression into an inx one.assume that all the numbers are single digit and unsigned .For simplicity, put all operation between parentheses. For instance, the postx expression ”23+” is transformed to ”(2+3)”, and ”873-*4+23-*58-+” is transformed to ”((((8*(73))+4)*(2-3))+(5-8))” (java)
// Private recursive method.
private static String recInfix(...) { ... }
// Public non-recursive
public static String infix(String exp) { ... }
Explanation / Answer
import java.util.*; public class Postfix { public static void main(String[] args) { System.out.println(postfixEvaluate("1 2 +")); // 3 System.out.println(postfixEvaluate("1 2 3 * + 4 +")); // 11 System.out.println(postfixEvaluate("5 2 4 * + 7 -")); // 6 System.out.println(postfixEvaluate("2 3 + 4 5 * +")); // 25 System.out.println(postfixEvaluate("8 5 * 7 4 2 + * +")); // 82 System.out.println(postfixEvaluate("6 8 2 / 1 - *")); // 18 } // Evaluates the given postfix expression string and returns the result. // Precondition: expression consists only of integers, +-*/, and spaces in // proper postfix format such as "2 3 - 4 *" public static int postfixEvaluate(String exp) { Stack s = new Stack (); Scanner tokens = new Scanner(exp); while (tokens.hasNext()) { if (tokens.hasNextInt()) { s.push(tokens.nextInt()); } else { int num2 = s.pop(); int num1 = s.pop(); String op = tokens.next(); if (op.equals("+")) { s.push(num1 + num2); } else if (op.equals("-")) { s.push(num1 - num2); } else if (op.equals("*")) { s.push(num1 * num2); } else { s.push(num1 / num2); } // "+", "-", "*", "/" } } return s.pop(); } }