<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.length()<num2.length()) multiply(num2,num1);
        vector <int> ans(num1.length()+num2.length(),0);int pro,carr=0;
        string ret;
        for(int i=num2.length()-1;i>=0;i--)
        {
            int w=ans.size()-(num2.length()-i);
            carr=0;
            for(int j=num1.length()-1;j>=0;j--)
            {
                pro=carr+(num2[i]-'0')*(num1[j]-'0');
                ans[w]+=pro%10;
                
                carr=pro/10;
                carr+=ans[w]/10;
                ans[w]%=10;
                w--;
            }
            if(carr!=0)
            {
                ans[w]+=carr;
            }
        }
        int flag=0;
        for(int i=0;i<ans.size();i++) {
            if(ans[i]!=0) flag++;
            if(ans[i]==0 && flag==0) continue;
            ret.push_back(ans[i]+'0');
        }
        if(ret.length()==0) return "0";
        return ret;
    }
};
8980