原题网址：https://leetcode.com/problems/add-digits/
Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.
For example:
Given num = 38, the process is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
Follow up:
Could you do it without any loop/recursion in O(1) runtime?
Hint:
- 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 Wikipedia article useful.
方法一：通过模拟数位相加的过程（循环或者递归）计算结果。
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11public class Solution { public int addDigits(int num) { if (num < 10) return num; int sum = 0; while (num > 0) { sum += num % 10; num /= 10; } return addDigits(sum); } }
方法二：通过数据样例分析找出规律。吓晕了，这是在提示下做出来的，可见分析简单数据样本找规律的重要性！
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6public class Solution { public int addDigits(int num) { if (num == 0) return 0; return (num-1)%9+1; } }
最后
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