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?
不用循环和递归!这个问题有公式可用,对于数字num,result=1+(num-1)%9。
可以参考wiki的Digital root。
1 class Solution { 2 public: 3 int addDigits(int num) { 4 int result=1+(num-1)%9; 5 return result; 6 } 7 };