150. Evaluate Reverse Polish Notation

1. Description

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.

2. Example

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

3. Constraints

• 1 <= tokens.length <= 10$^{4}$
• tokens[i] is either an operator: “+”, “-”, “*”, or “/”, or an integer in the range [-200, 200].

4. Solutions

Stack

n = tokens.size()
Time complexity: O(n)
Space complexity: O(n)

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class Solution {
public:
    int evalRPN(vector<string> &tokens) {
        stack<int> numbers;
        for (const string &token : tokens) {
            if (token.size() > 1 || token[0] >= '0') {
                numbers.push(stoi(token));
            } else {
                int num2 = numbers.top();
                numbers.pop();
                int num1 = numbers.top();
                numbers.pop();

                switch (token[0]) {
                case '+':
                    numbers.push(num1 + num2);
                    break;
                case '-':
                    numbers.push(num1 - num2);
                    break;
                case '*':
                    numbers.push(num1 * num2);
                    break;
                case '/':
                    numbers.push(num1 / num2); 
                    break;
                }
            }
        }

        return numbers.top();
    }
};