題目

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

Example:

Input: 38
Output: 2 
Explanation: 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?

十分鐘嘗試

利用遞迴呼叫，除以10，對10求mod獲取每位數字，邏輯不是問題，已經一遍ok，程式碼如下，但是題目的followup需要研究一下，先上一個遞迴版本：

class Solution {
    public int addDigits(int num) {
        int res=caculate(num);
        if(res/10==0){
            return res;
        }
        return addDigits(res);
    }
    private int caculate(int num){
        int res=0;
        while(num>0){
            res+=num%10;
           num=num/10;   
        }
        return res;
        
    }
}
 

非遞迴非迴圈解法

class Solution {
    public int addDigits(int num) {
       return 1 + (num - 1) % 9;
    }
   
}