<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) {
        vector<int> n1;
        vector<int> n2;
        

        for(int i=num1.length()-1;i>=0;i--){
            n1.push_back(num1[i]-'0');
        }
        for(int i=num2.length()-1;i>=0;i--){
            n2.push_back(num2[i]-'0');
        }

        int a=n1.size();
        int b=n2.size();
        vector<int> ans(a+b);

        for(int i=0;i<a;i++){
            for(int j=0;j<b;j++){
                ans[i+j]+=n1[i]*n2[j];
            }
        }

        int add=0;
        string result="";
        for(int i=0;i<ans.size();i++){
            ans[i]+=add;
            add=ans[i]/10;
            result=to_string(ans[i]%10)+result;
        }
        result=to_string(add)+result;
        while(result[0]=='0') result.erase(0,1);

        if(result.empty()) return "0";

        return result;
    }
};
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