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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?

题意：把2一个数拆开，最终变成一个1-9的数字

最先看到题先想到的是递归拆，勉强过了

Java

 public int addDigits(int num) {
        System.out.println(num);
        if (num <= 9)
            return num;
        int sum = 0;
        while (num != 0) {
            sum += num%10;
            num = num/10;
        }
        return addDigits(sum);
    }
}

显然题意不可能是用这个算法，末尾也有提示说O(1) 的算法复杂度，那就只能找规律了

1-9   10 11 12 13 18  19-27

1-9   1-9                       1-9

规律就比较明显了

Java

class Solution {

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