<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) {
        if (num1 == "0" || num2 == "0") {
            return "0";
        }
        int m = num1.size(), n = num2.size();
        vector<int> arr(m + n);
        for (int i = m - 1; i >= 0; --i) {
            int a = num1[i] - '0';
            for (int j = n - 1; j >= 0; --j) {
                int b = num2[j] - '0';
                arr[i + j + 1] += a * b;
            }
        }
        for (int i = arr.size() - 1; i; --i) {
            arr[i - 1] += arr[i] / 10;
            arr[i] %= 10;
        }
        int i = arr[0] ? 0 : 1;
        string ans;
        for (; i < arr.size(); ++i) {
            ans += '0' + arr[i];
        }
        return ans;
    }
};
        
   

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