LeetCode 257: Add Digits | CrazyEgg

LeetCode: Add Digits | CrazyEgg
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?

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.

    public int addDigits(int num) {
        while (num > 9) {
            num = getInt(num);
        }
        return num;
    }

    private int getInt(int num) {
        int result = 0;
        while (num >= 10) {
            result += num % 10;
            num /= 10;
        }
        result += num;
        return result;
    }

http://www.cnblogs.com/maples7/p/4734300.html
http://shibaili.blogspot.com/2015/08/day-121-264-ugly-number-ii.html

    int addDigits(int num) {

        if (num == 0) return 0;

        return num - (num - 1) / 9 * 9;

    }

http://blog.csdn.net/ER_Plough/article/details/48157351

给定一个整数，计算各位上的数字的和得到另外一个数字，直到该数字为个位数。

模拟一下就可以得出结果，但是要求o(1)的时间复杂度，找规律！

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

http://www.cnblogs.com/ganganloveu/p/4749964.html

解法二、套公式digital root

    int addDigits(int num) {
        return num - 9 * floor((num - 1) / 9);
    }

Read full article from LeetCode: Add Digits | CrazyEgg