<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>

class Solution {
public:
    string multiply(string num1, string num2) {
        int s1 = num1.size();
        int s2 = num2.size();
        vector<int> values(s1 + s2 + 1);
        for (int i = 0; i < s1; ++i) {
            for (int j = 0 ; j < s2; ++j) {
                int cur = (num1.at(s1 - 1 - i) - '0') * (num2.at(s2 - 1 - j) - '0');
                values[i + j + 1] += cur / 10;
                values[i + j] += cur % 10;
            }
        }
        for (int i = 0; i < values.size() - 1; ++i) {
            int cur = values[i];
            values[i] = cur % 10;
            values[i + 1] += cur / 10;
        }
        int index = values.size() - 1;
        string res = "";
        while(index >= 0 && values[index] == 0) {
            index--;
        }
        while(index >= 0) {
            res+= to_string(values[index--]);
        }
        if (res.empty()) {
            return "0";
        }
        return res;
    }
};
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