<p>Given an integer <code>num</code>, repeatedly add all its digits until the result has only one digit, and return it.</p>

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

<pre>
<strong>Input:</strong> num = 38
<strong>Output:</strong> 2
<strong>Explanation:</strong> The process is
38 --&gt; 3 + 8 --&gt; 11
11 --&gt; 1 + 1 --&gt; 2 
Since 2 has only one digit, return it.
</pre>

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

<pre>
<strong>Input:</strong> num = 0
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>0 &lt;= num &lt;= 2<sup>31</sup> - 1</code></li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you do it without any loop/recursion in <code>O(1)</code> runtime?</p>

class Solution:
    def addDigits(self, num: int) -> int:
        s = 0
        while num != 0:
            s += num % 10
            num = int(num // 10)
        if len(str(s)) != 1:
            return self.addDigits(s)
        else:
            return s
        
        
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