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):

 1 public class Solution {
 2     public int addDigits(int num) {
 3         if (num == 0)
 4             return 0 ;
 5         if (num > 9)
 6             num %= 9 ;
 7         if (num == 0)
 8             num = 9 ; 
 9         return num;
10     }
11 }

C++(6ms):

 1 class Solution {
 2 public:
 3     int addDigits(int num) {
 4         if (num == 0)
 5           return 0 ;
 6         if(num%9 == 0)
 7           return 9 ;
 8         else
 9           return num%9 ;
10     }
11 };