<p>Given two non-negative integers, <code>num1</code> and <code>num2</code> represented as string, return <em>the sum of</em> <code>num1</code> <em>and</em> <code>num2</code> <em>as a string</em>.</p>

<p>You must solve the problem without using any built-in library for handling large integers (such as <code>BigInteger</code>). You must also not convert the inputs to integers directly.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> num1 = &quot;11&quot;, num2 = &quot;123&quot;
<strong>Output:</strong> &quot;134&quot;
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> num1 = &quot;456&quot;, num2 = &quot;77&quot;
<strong>Output:</strong> &quot;533&quot;
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> num1 = &quot;0&quot;, num2 = &quot;0&quot;
<strong>Output:</strong> &quot;0&quot;
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= num1.length, num2.length &lt;= 10<sup>4</sup></code></li>
	<li><code>num1</code> and <code>num2</code> consist of only digits.</li>
	<li><code>num1</code> and <code>num2</code> don&#39;t have any leading zeros except for the zero itself.</li>
</ul>

class Solution {
public:
    string addStrings(string num1, string num2) {
        int n1 = num1.size();
        int n2 = num2.size();
        int n = max(n1, n2);
        
        for (int i = n1; i < n; i++) {
            num1 = "0" + num1;
        }
        
        for (int i = n2; i < n; i++) {
            num2 = "0" + num2;
        }
        
        string ans = "";
        int carry = 0;
        char c;
        for (int i = n - 1; i >= 0; i--) {
            int d1 = num1[i] - '0';
            int d2 = num2[i] - '0';
            int d = d1 + d2 + carry;
            carry = d / 10;
            d = d % 10;
            c = ('0' + d);
            ans = c + ans;
        }
        
        c = ('0' + carry);
        if (carry) ans = c + ans;
        
        return ans;
    }
};
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