<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 {
    vector<int> stringToVec(const string& num) {
        vector<int> v(num.size());
        for (int i = 0; i < num.size(); ++i) {
            v[i] = num[i] - '0';
        }
        return v;
    }
public:
    string addStrings(string num1, string num2) {
        vector<int> v1 = stringToVec(num1);
        vector<int> v2 = stringToVec(num2);
        int i = v1.size() - 1, j = v2.size() - 1;
        int c = 0;
        string s = "";
        while (i >= 0 && j >= 0) {
            int n = v1[i] + v2[j] + c;
            char d = (n % 10) + '0';
            s = d + s;
            c = n / 10;
            --i;
            --j;
        }
        while (i >= 0) {
            int n = v1[i] + c;
            char d = (n % 10) + '0';
            s = d + s;
            c = n / 10;
            --i;
        }
        while (j >= 0) {
            int n = v2[j] + c;
            char d = (n % 10) + '0';
            s = d + s;
            c = n / 10;
            --j;
        }
        if (c != 0) {
            char d = c + '0';
            s = d + s;
        }
        return s;
    }
};
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