LeetCode: 258 Add Digits(easy)

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?

代码：

 class Solution {
 public:
     int addDigits(int num) {
         return !num ?  : (num - ) %  + ;
     }
 };

思路：

The problem, widely known as digit root problem, has a congruence formula:

https://en.wikipedia.org/wiki/Digital_root#Congruence_formula
For base b (decimal case b = 10), the digit root of an integer is:

dr(n) = 0 if n == 0
dr(n) = (b-1) if n != 0 and n % (b-1) == 0
dr(n) = n mod (b-1) if n % (b-1) != 0
or

dr(n) = 1 + (n - 1) % 9
Note here, when n = 0, since (n - 1) % 9 = -1, the return value is zero (correct).

From the formula, we can find that the result of this problem is immanently periodic, with period (b-1).

Output sequence for decimals (b = 10):

~input: 0 1 2 3 4 ...
output: 0 1 2 3 4 5 6 7 8 9 1 2 3 4 5 6 7 8 9 1 2 3 ....

Henceforth, we can write the following code, whose time and space complexities are both O(1).

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