258. Add Digits

题目：

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.

链接： http://leetcode.com/problems/add-digits/

题解：

又是数学题，求digital root。循环叠加比较容易，但看了wiki以后发现了公式，还是用公式算吧。这种数学题对数学不好的我来说真是头大。原理10 % 9 或者 100 % 9都等于 1 % 9。举个例子n = abc = a  * 100 + b * 10 + c，那么 (a*100 + b * 10 + c) % 9 = (a + b + c) % 9。由此n == 0时，result = 0， n % 9 == 0时， 说明a + b + c = 9，我们返回9，对于其他数字， (a + b + c)等于res % 9。  

Time Complexity - O(1)， Space Complexity - O(1)

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

Reference:

https://en.wikipedia.org/wiki/Digital_root

https://leetcode.com/discuss/67755/3-methods-for-python-with-explains

https://leetcode.com/discuss/52122/accepted-time-space-line-solution-with-detail-explanations

https://leetcode.com/discuss/55910/two-lines-c-code-with-explanation

258. Add Digits

原文：http://www.cnblogs.com/yrbbest/p/5014963.html