<p>Given a string <code>s</code> representing a valid expression, implement a basic calculator to evaluate it, and return <em>the result of the evaluation</em>.</p>

<p><strong>Note:</strong> You are <strong>not</strong> allowed to use any built-in function which evaluates strings as mathematical expressions, such as <code>eval()</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;1 + 1&quot;
<strong>Output:</strong> 2
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> s = &quot; 2-1 + 2 &quot;
<strong>Output:</strong> 3
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> s = &quot;(1+(4+5+2)-3)+(6+8)&quot;
<strong>Output:</strong> 23
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= s.length &lt;= 3 * 10<sup>5</sup></code></li>
	<li><code>s</code> consists of digits, <code>&#39;+&#39;</code>, <code>&#39;-&#39;</code>, <code>&#39;(&#39;</code>, <code>&#39;)&#39;</code>, and <code>&#39; &#39;</code>.</li>
	<li><code>s</code> represents a valid expression.</li>
	<li><code>&#39;+&#39;</code> is <strong>not</strong> used as a unary operation (i.e., <code>&quot;+1&quot;</code> and <code>&quot;+(2 + 3)&quot;</code> is invalid).</li>
	<li><code>&#39;-&#39;</code> could be used as a unary operation (i.e., <code>&quot;-1&quot;</code> and <code>&quot;-(2 + 3)&quot;</code> is valid).</li>
	<li>There will be no two consecutive operators in the input.</li>
	<li>Every number and running calculation will fit in a signed 32-bit integer.</li>
</ul>

class Solution {
public:
    int calculate(string s) {
        if (s.size() == 0) {
            return 0;
        }
        vector<int> st;
        char sign = '+';
        long number = 0;
        
        for (int i = 0; i < s.size(); i++) {
            if (isdigit(s[i])) {
                number = number * 10 + (long)(s[i] - '0');
            } else if (s[i] == '(') {
                int braces = 1;
                int j = i + 1;
                while (braces > 0) {
                    if (s[j] == '(') {
                        braces++;
                    } else if (s[j] == ')') {
                        braces--;
                    }
                    j++;
                }
                int length = (j - 1) - i;
                number = calculate(s.substr(i + 1, length));
                i = j - 1;
            }
            
            if (s[i] == '+' || s[i] == '-' || i == s.size() - 1) {
                switch(sign) {
                    case '+': {
                        st.push_back(number);
                        break;
                    }
                    case '-': {
                        st.push_back(-number);
                        break;
                    }
                }
                sign = s[i];
                number = 0;
            }
        }
        int sum = 0;
        while (!st.empty()) {
            sum += st.back();
            st.pop_back();
        }
        return sum;
    }
};
78743