Title

the problem link

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?

Credits: Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

Solution

Digital Root

this method depends on the truth:

N=(a[0] * 1 + a[1] * 10 + …a[n] * 10 ^n),and a[0]…a[n] are all between [0,9]

we set M = a[0] + a[1] + …a[n]

and another truth is that:

1 % 9 = 1

10 % 9 = 1

100 % 9 = 1

so N % 9 = a[0] + a[1] + …a[n]

means N % 9 = M

so N = M (% 9)

as 9 % 9 = 0,so we can make (n - 1) % 9 + 1 to help us solve the problem when n is 9.as N is 9, ( 9 - 1) % 9 + 1 = 9

//cpp:
class Solution {
public:
    int addDigits(int num) {

        int res = num % 9;
        return (res != 0 || num == 0) ? res : 9;

    }
};