<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:
/*
    858
     45
   4290   carry  0 
  34220
  38510

*/
    string multiply(string num1, string num2) {
        if (num1 == "0" || num2 == "0") {
            return "0";
        }
        if (num1.length() < num2.length()) {
            string temp = num1;
            num1 = num2; // first operand
            num2 = temp; // multiplyer
        }
        // Reverse sring
        reverse(num1.begin(), num1.end());
        reverse(num2.begin(), num2.end());

        // Allocate an array for addition initialised with 0
        vector<int> answer(2*max(num1.length(), num2.length()), 0);
        for (int i = 0; i < num2.length(); ++i) {
            vector<int> multiplication = multiplyDigit(num1, num2[i], i);
            answer = addDigit(multiplication, answer);
        }

        // Remove extra 0 at the end
        while (answer.back() == 0) {
            answer.pop_back();
        }

        // Put together the result
        string result = "";
        for (int i = answer.size()-1; i >= 0; i--) {
            result += answer[i] + '0'; // To make it a char
        }

        return result;
    }

    vector<int> multiplyDigit(string num1, char digit, int order) {
        vector<int> multiplication(order, 0);
        int carry = 0;
        for (int i = 0; i < num1.length(); ++i) {
            int ret = (num1[i] - '0') * (digit - '0') + carry;
            multiplication.push_back(ret % 10);
            carry = ret/10;
        }
        if (carry > 0) {
            multiplication.push_back(carry);
        }
        return multiplication;
    }

    vector<int> addDigit(vector<int> op1, vector<int> op2) {
        vector<int> addition;
        int carry = 0;
        for (int i = 0; i < max(op1.size(), op2.size()) || carry != 0; ++i) {
            int d1 = (i < op1.size()) ? op1[i] : 0;
            int d2 = (i < op2.size()) ? op2[i] : 0;
            int ret = d1 + d2 + carry;
            addition.push_back(ret % 10);
            carry = ret/10;
        }
        return addition;
    }
};
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