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            "categoryTitle": "Algorithms",
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            "title": "Add Digits",
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            "content": "<p>Given an integer <code>num</code>, repeatedly add all its digits until the result has only one digit, and return it.</p>\n\n<p>&nbsp;</p>\n<p><strong>Example 1:</strong></p>\n\n<pre>\n<strong>Input:</strong> num = 38\n<strong>Output:</strong> 2\n<strong>Explanation:</strong> The process is\n38 --&gt; 3 + 8 --&gt; 11\n11 --&gt; 1 + 1 --&gt; 2 \nSince 2 has only one digit, return it.\n</pre>\n\n<p><strong>Example 2:</strong></p>\n\n<pre>\n<strong>Input:</strong> num = 0\n<strong>Output:</strong> 0\n</pre>\n\n<p>&nbsp;</p>\n<p><strong>Constraints:</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= num &lt;= 2<sup>31</sup> - 1</code></li>\n</ul>\n\n<p>&nbsp;</p>\n<p><strong>Follow up:</strong> Could you do it without any loop/recursion in <code>O(1)</code> runtime?</p>\n",
            "translatedTitle": "各位相加",
            "translatedContent": "<p>给定一个非负整数 <code>num</code>，反复将各个位上的数字相加，直到结果为一位数。返回这个结果。</p>\n\n<p>&nbsp;</p>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入:</strong> num =<strong> </strong><code>38</code>\n<strong>输出:</strong> 2 \n<strong>解释: </strong>各位相加的过程为<strong>：\n</strong>38 --&gt; 3 + 8 --&gt; 11\n11 --&gt; 1 + 1 --&gt; 2\n由于&nbsp;<code>2</code> 是一位数，所以返回 2。\n</pre>\n\n<p><strong>示例 1:</strong></p>\n\n<pre>\n<strong>输入:</strong> num =<strong> </strong>0\n<strong>输出:</strong> 0</pre>\n\n<p>&nbsp;</p>\n\n<p><strong>提示：</strong></p>\n\n<ul>\n\t<li><code>0 &lt;= num &lt;= 2<sup>31</sup>&nbsp;- 1</code></li>\n</ul>\n\n<p>&nbsp;</p>\n\n<p><strong>进阶：</strong>你可以不使用循环或者递归，在 <code>O(1)</code> 时间复杂度内解决这个问题吗？</p>\n",
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                "A naive implementation of the above process is trivial. Could you come up with other methods?",
                "What are all the possible results?",
                "How do they occur, periodically or randomly?",
                "You may find this <a href=\"https://en.wikipedia.org/wiki/Digital_root\" target=\"_blank\">Wikipedia article</a> useful."
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