## [224\. Basic Calculator](https://leetcode.com/problems/basic-calculator/)

Given a string `s` representing a valid expression, implement a basic calculator to evaluate it, and return _the result of the evaluation_.

**Note:** You are **not** allowed to use any built-in function which evaluates strings as mathematical expressions, such as `eval()`.

**Example 1:**

**Input:** s = "1 + 1"
**Output:** 2

**Example 2:**

**Input:** s = " 2-1 + 2 "
**Output:** 3

**Example 3:**

**Input:** s = "(1+(4+5+2)-3)+(6+8)"
**Output:** 23

**Constraints:**

- `1 <= s.length <= 3 * 105`
- `s` consists of digits, `'+'`, `'-'`, `'('`, `')'`, and `' '`.
- `s` represents a valid expression.
- `'+'` is **not** used as a unary operation (i.e., `"+1"` and `"+(2 + 3)"` is invalid).
- `'-'` could be used as a unary operation (i.e., `"-1"` and `"-(2 + 3)"` is valid).
- There will be no two consecutive operators in the input.
- Every number and running calculation will fit in a signed 32-bit integer.

class Solution {
public:
    int calculate(string s) {
        stack<int> stk;
        int res = 0;
        int sign = 1;
        int operand = 0;
        for(int i = 0; i < s.size(); i++) {
            char c = s[i];
            if(isdigit(c)) {
                operand = 10 * operand + (c - '0');
            } else if(c == '+') {
                res += sign * operand;
                sign = 1;
                // 已經將答案存到 res 了，operand reset 為 0
                operand = 0;
            } else if(c == '-') {
                res += sign * operand;
                sign = -1;
                // 已經將答案存到 res 了，operand reset 為 0
                operand = 0;
            } else if(c == '(') {
                // 將結果暫時存到 stack
                stk.push(res);

                // 也將 sign 暫時存到 stack
                stk.push(sign);

                // reset 結果為 0，重新計算
                res = 0;
                
                // reset sign 為 1，重新計算
                sign = 1;
            } else if(c == ')') {
                // 加上目前的計算
                res += sign * operand;

                // 乘上之前運算的結果的 sign
                res *= stk.top(); stk.pop();

                // 加上之前運算的結果
                res += stk.top(); stk.pop();

                // reset operand 為 0
                operand = 0;
            }
        }
        return res + (sign * operand);
    }
};

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