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?
输入：1,2,3,4,5,6,7,8,9,10 11, 12, 13, 14, 15, 16， 17。 18， 19，20
结果：1, 2,3,4,5,6,7,8,9,1, 2, 3, 4, 5, 6, 。7， 8， 9， 1， 2依据上面的尝试，我们能够发现，这里是有规律的。
即结果满足，（num-1）%9+1
public static int addDigits(int num){
return (num-1)%9+1;
}