LeetCode 258. Add Digits（数位相加）

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:

1. A naive implementation of the above process is trivial. Could you come up with other methods?
2. What are all the possible results?
3. How do they occur, periodically or randomly?
4. 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。

public 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！

public class Solution {    public int addDigits(int num) {        if (num == 0) return 0;        return (num-1)%9+1;    }}