Given an integer num, repeatedly add all its digits until the result has only one digit, and return it.
\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\n0 <= num <= 231 - 1\n
Follow up: Could you do it without any loop/recursion in O(1) runtime?
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
示例 1:
\n\n\n输入: num =\n\n38\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
示例 2:
\n\n\n输入: num = 0\n输出: 0\n\n
\n\n
提示：
\n\n0 <= num <= 231 - 1\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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