Multiply Strings
<p>Given two non-negative integers <code>num1</code> and <code>num2</code> represented as strings, return the product of <code>num1</code> and <code>num2</code>, also represented as a string.</p>

<p><strong>Note:</strong>&nbsp;You must not use any built-in BigInteger library or convert the inputs to integer directly.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<pre><strong>Input:</strong> num1 = "2", num2 = "3"
<strong>Output:</strong> "6"
</pre><p><strong class="example">Example 2:</strong></p>
<pre><strong>Input:</strong> num1 = "123", num2 = "456"
<strong>Output:</strong> "56088"
</pre>
<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num1.length, num2.length &lt;= 200</code></li>
	<li><code>num1</code> and <code>num2</code> consist of digits only.</li>
	<li>Both <code>num1</code> and <code>num2</code>&nbsp;do not contain any leading zero, except the number <code>0</code> itself.</li>
</ul>
Math; String; Simulation
impl Solution {
    pub fn multiply(num1: String, num2: String) -> String {
        if num1 == "0" || num2 == "0" {
            return String::from("0");
        }
        let (num1, num2) = (num1.as_bytes(), num2.as_bytes());
        let (n, m) = (num1.len(), num2.len());
        let mut res = vec![];
        for i in 0..n {
            let a = num1[n - i - 1] - b'0';
            let mut sum = 0;
            let mut j = 0;
            while j < m || sum != 0 {
                if i + j == res.len() {
                    res.push(0);
                }
                let b = num2.get(m - j - 1).unwrap_or(&b'0') - b'0';
                sum += a * b + res[i + j];
                res[i + j] = sum % 10;
                sum /= 10;
                j += 1;
            }
        }
        res.into_iter()
            .rev()
            .map(|v| char::from(v + b'0'))
            .collect()
    }
}
