{ "data": { "question": { "questionId": "258", "questionFrontendId": "258", "categoryTitle": "Algorithms", "boundTopicId": 1117, "title": "Add Digits", "titleSlug": "add-digits", "content": "

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

\n\n

 

\n

Example 1:

\n\n
\nInput: num = 38\nOutput: 2\nExplanation: The process is\n38 --> 3 + 8 --> 11\n11 --> 1 + 1 --> 2 \nSince 2 has only one digit, return it.\n
\n\n

Example 2:

\n\n
\nInput: num = 0\nOutput: 0\n
\n\n

 

\n

Constraints:

\n\n\n\n

 

\n

Follow up: Could you do it without any loop/recursion in O(1) runtime?

\n", "translatedTitle": "各位相加", "translatedContent": "

给定一个非负整数 num，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。

\n\n

 

\n\n

示例 1:

\n\n
\n输入: num = 38\n输出: 2 \n解释: 各位相加的过程为：\n38 --> 3 + 8 --> 11\n11 --> 1 + 1 --> 2\n由于 2 是一位数，所以返回 2。\n
\n\n

示例 1:

\n\n
\n输入: num = 0\n输出: 0
\n\n

 

\n\n

提示：

\n\n\n\n

 

\n\n

进阶：你可以不使用循环或者递归，在 O(1) 时间复杂度内解决这个问题吗？

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