(leetcode)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?

 class Solution {
 public:
     int addDigits(int num) {
         int ans = num;
         // while(ans!=2)
         // {
         while()
         {
             if(ans < ) return ans;
             num = ans;
             ans = ;
             while(num != )
             {
                 ans += num%;
                 num /= ;
             }
         }
     }
 };

O(1)方法

观察如下输入：

输入：1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20

输出：1,2,3,4,5,6,7,8,9,1 , 2 , 3 , 4 , 5 , 6, 7 , 8 , 9 , 1 , 2

可以看到返回的值就是(num-1)%9+1