Basic Calculator
<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>
Stack; Recursion; Math; String
function calculate(s: string): number {
    const stk: number[] = [];
    let sign = 1;
    let ans = 0;
    const n = s.length;
    for (let i = 0; i < n; ++i) {
        if (s[i] === ' ') {
            continue;
        }
        if (s[i] === '+') {
            sign = 1;
        } else if (s[i] === '-') {
            sign = -1;
        } else if (s[i] === '(') {
            stk.push(ans);
            stk.push(sign);
            ans = 0;
            sign = 1;
        } else if (s[i] === ')') {
            ans *= stk.pop() as number;
            ans += stk.pop() as number;
        } else {
            let x = 0;
            let j = i;
            for (; j < n && !isNaN(Number(s[j])) && s[j] !== ' '; ++j) {
                x = x * 10 + (s[j].charCodeAt(0) - '0'.charCodeAt(0));
            }
            ans += sign * x;
            i = j - 1;
        }
    }
    return ans;
}
