LeetCode题解——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?

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;    }};

Thanks for reading. :)