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?

Hint:

1. A naive implementation of the above process is trivial. Could you come up with other methods?
2. What are all the possible results?
3. How do they occur, periodically or randomly?
4. You may find this Wikipedia article useful.

解题思路

思路1：一直相加直到num变为个位数。
思路2：套用维基公式。

实现代码

Java代码 1：

// Runtime: 2 ms
public class Solution {
    public int addDigits(int num) {
        int temp = 0;
        while (num > 9) {
            while (num != 0) {
                temp += num % 10;
                num /= 10;
            }
            num = temp;
            temp = 0;
        }

        return num;
    }
}

Java代码2：

// Runtime: 5 ms
public class Solution {
    public int addDigits(int num) {
        return (int)(num - 9 * Math.ceil((num - 1) / 9));
    }
}

Java代码3：

// Runtime: 2 ms
public class Solution {
    public int addDigits(int num) {
        return 1 + (num - 1) % 9;
    }
}