<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& A, string& B) {
        if(A == "0" || B == "0") return "0";
        string ans(size(A)+size(B), '0');
        for(int i = size(A)-1; i >= 0; i--) {
            for(int j = size(B)-1; j >= 0; j--) {
                int res = (ans[i+j+1]-'0') + (A[i]-'0') * (B[j]-'0');
                ans[i+j+1] = res%10 + '0';  // ones place of multiplication result
                ans[i+j] += res/10;         // carry over to previous index if result >= 10
            }
        }
        if(ans[0] == '0') return ans.substr(1);
        return ans;
    }
};
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