258. 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?

不用循环

 xyz = 100x + 10y +z, xyz % 9 = 99*x + 9*y + (x + y + z) % 9 = (x + y + z) % 9

java(3ms):

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

C++(6ms):

class Solution {
public:
    int addDigits(int num) {
        if (num == 0)
          return 0 ;
        if(num%9 == 0)
          return 9 ;
        else
          return num%9 ;
    }
};