## [150\. Evaluate Reverse Polish Notation](https://leetcode.com/problems/evaluate-reverse-polish-notation/)

You are given an array of strings `tokens` that represents an arithmetic expression in a [Reverse Polish Notation](http://en.wikipedia.org/wiki/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]`.

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<int> st;
        for(auto& token : tokens) {
            if(token != "+" && token != "-" && token != "*" && token != "/") {
                st.push(stoi(token));
            } else {
                int num1 = st.top(); st.pop();
                int num2 = st.top(); st.pop();
                if(token == "+") st.push(num2 + num1);
                if(token == "-") st.push(num2 + num1);
                if(token == "*") st.push(num2 * num1);
                if(token == "/") st.push(num2 / num1);
            }
        }
        return st.top();
    }
};

:::success
- 時間複雜度：$O(n)$
- 空間複雜度：$O(n)$
:::