<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;
        int tot = 0;
        int op = 0;
        int i = 0;
        while (i < s.size()) {
            int rhs = 0;
            while (s[i] == ' ') {
                i++;
            }

            if (s[i] == '+') {
                op = 1;
                i++;
            } else if (s[i] == '-') {
                op = -1;
                i++;
            }

            while (s[i] >= '0' && s[i] <= '9') {
                rhs *= 10;
                rhs += s[i] - '0';
                i++;
            }

            if (s[i] == '(') {
                int parens = 1;
                i++;
                int start = i;
                while (parens > 0) {
                    switch (s[i]) {
                        case '(':
                            parens++;
                            break;
                        case ')':
                            parens--;
                    }
                    i++;
                }

                rhs = calculate(s.substr(start, i - start - 1));
            }

            while (s[i] == ' ') {
                i++;
            }

            if (op == 0)
                tot = rhs;
            else
                tot += rhs * op;
        }

        return tot;
    }
};
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