Evaluate a Postfix Expression

Problem

A postfix expression has numbers and the operators + - * / separated by spaces; the operator comes after its two numbers (3 4 + means 3 + 4). Print the value. Division cuts the result toward zero. The input is always valid and never divides by zero.

Constraints

  • 1 <= tokens <= 10^4
  • |intermediate values| <= 10^9

Examples

Input

5 1 2 + 4 * + 3 -

Output

14

Solutions

Best approachTime O(n)Space O(n)

Postfix needs no brackets and no priority rules; one left-to-right pass with a stack is enough. Java and JavaScript cut division toward zero; Python's // rounds down, so we divide the absolute values and fix the sign.

Same code works in Java810172125Basic features only, so nothing changes between versions.
/*
 * Problem : Postfix value.
 * Input   : One line of tokens.
 * Output  : The value.
 * Approach: Use a stack.
 * Time    : O(n)
 * Space   : O(n)
 * Java    : 8 or newer.
 */
import java.util.Scanner;

public class Main {
    // This method solves the task. It receives the input and returns the answer.
    static long solve(String s) {
        // Step 1: A stack holds the numbers that wait for an operator.
        java.util.ArrayDeque<Long> stack = new java.util.ArrayDeque<>();
        for (String token : s.trim().split("\\s+")) {
            if (token.length() == 1 && "+-*/".contains(token)) {
                // Step 2: An operator takes the two top numbers. The second number from the top is the left side.
                long right = stack.pop();
                long left = stack.pop();
                if (token.equals("+")) {
                    stack.push(left + right);
                } else if (token.equals("-")) {
                    stack.push(left - right);
                } else if (token.equals("*")) {
                    stack.push(left * right);
                } else {
                    stack.push(left / right);
                }
            } else {
                stack.push(Long.parseLong(token));
            }
        }
        // Step 3: At the end, one number is left. It is the answer.
        return stack.pop();
    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        String s = scanner.hasNextLine() ? scanner.nextLine() : "";
        System.out.println(solve(s));
    }
}
Hints (1)
  1. Why is the right operand popped first?