<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:
/*
expr := 0 | [1-9][0-9]*
expr := (expr)
expr := expr + expr
expr := expr - expr
*/
    int calculate(string s) {
        int acc = 0;
        char op = '+';

        for(int i = 0; i < s.size(); ++i) {
            if(s[i] == ' ') continue; // whitespace
            else if(isdigit(s[i])) { // number
                int len = 1;
                while(i+len < s.size()) {
                    if(!isdigit(s[i+len])) break;
                    len++;
                }
                int n = stoi(s.substr(i,len));
                acc += (op == '+' ? n : -n);
                i += len-1;
            }
            else if(s[i] == '+' || s[i] == '-') { 
                op = s[i];
            }
            else if(s[i] == '(') {
                int bcount = 1;
                int len = 0;
                while(bcount != 0 && i+len < s.size()) {
                    ++len;
                    if(s[i+len] == ')') --bcount;
                    else if(s[i+len] == '(') ++bcount;
                }

                int n = calculate(s.substr(i+1, len-1));
                
                acc += (op == '+' ? n : -n);
                i += len;
            }
        }
        
        return acc;
    }
};
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