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150. Evaluate Reverse Polish Notation

You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation.

Evaluate the expression. Return an integer that represents the value of the expression.

Note that:

  • The valid operators are '+', '-', '*', and '/'.
  • Each operand may be an integer or another expression.
  • The division between two integers always truncates toward zero.
  • There will not be any division by zero.
  • The input represents a valid arithmetic expression in a reverse polish notation.
  • The answer and all the intermediate calculations can be represented in a 32-bit integer.

Example 1:

Input: tokens = ["2","1","+","3","*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

Example 2:

Input: tokens = ["4","13","5","/","+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6

Example 3:

Input: tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]
Output: 22
Explanation: ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

Constraints:

  • 1 <= tokens.length <= 104
  • tokens[i] is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].

Solution:

class Solution {
    public int evalRPN(String[] tokens) {
        Deque<Integer> stack = new ArrayDeque<Integer>();
        for (int i = 0; i < tokens.length; i++){
            if ( (!tokens[i].equals("+") ) && (!tokens[i].equals("-")) &&
            (!tokens[i].equals("/")) && (!tokens[i].equals("*"))){
                stack.offerLast(Integer.parseInt(tokens[i]));
            }else{
                int a = stack.pollLast();
                int b = stack.pollLast();
                int c = 0;

                if (tokens[i].equals("/")){
                    c = b / a;
                }else if (tokens[i].equals("+")){
                    c = b + a;
                }else if (tokens[i].equals("*")){
                    c = b * a;
                }else if (tokens[i].equals("-")){
                    c = b - a;
                }

                stack.offerLast(c);
            }
        }

        return stack.pollLast();
    }
}