<p>Given a string <code>s</code> which represents an expression, <em>evaluate this expression and return its value</em>.&nbsp;</p>

<p>The integer division should truncate toward zero.</p>

<p>You may assume that the given expression is always valid. All intermediate results will be in the range of <code>[-2<sup>31</sup>, 2<sup>31</sup> - 1]</code>.</p>

<p><strong>Note:</strong> You are not 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 = "3+2*2"
<strong>Output:</strong> 7
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> s = " 3/2 "
<strong>Output:</strong> 1
</pre><p><strong class="example">Example 3:</strong></p>
<pre><strong>Input:</strong> s = " 3+5 / 2 "
<strong>Output:</strong> 5
</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 integers and operators <code>(&#39;+&#39;, &#39;-&#39;, &#39;*&#39;, &#39;/&#39;)</code> separated by some number of spaces.</li>
	<li><code>s</code> represents <strong>a valid expression</strong>.</li>
	<li>All the integers in the expression are non-negative integers in the range <code>[0, 2<sup>31</sup> - 1]</code>.</li>
	<li>The answer is <strong>guaranteed</strong> to fit in a <strong>32-bit integer</strong>.</li>
</ul>

class Solution:
    def calculate(self, s: str) -> int:
        if not s:
            return 0
        stack = []
        cur = 0
        op = '+'
        for i in range(len(s)):
            if s[i].isdigit():
                cur = cur * 10 + int(s[i])
            # Process the number when we hit an operator or at the last character
            if s[i] in "+-*/" or i == len(s) - 1:
                if op == '+':
                    stack.append(cur)
                elif op == '-':
                    stack.append(-cur)
                elif op == '*':
                    stack.append(stack.pop() * cur)
                elif op == '/':
                    prev = stack.pop()
                    # Simulate division truncating toward zero
                    stack.append(int(prev / cur))
                op = s[i]
                cur = 0
        
        return sum(stack)

19800