leetcode 258: Add Digits

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题目：

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:

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

思路：

不用循环或者递归的时候，没有做出来；提示没看懂，也没有认真思考；

编码1：

public class Solution {
    public int addDigits(int num) {
        int sum = 0;
        while(true) {
            sum += num%10;
            num = num/10;
            if(num == 0) {
                if(sum < 10) 
                    break;
                else {
                    num = sum;
                    sum = 0;
                }
            }
        }
        return sum;
    }
}

编码2：

针对follow up，一共有多少种结果？1-9；每种结果是随机的还是有规律的？（注释：代码为粘贴copy其他人的）

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

leetcode 258: Add Digits

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原文地址：http://www.cnblogs.com/cactus1504/p/4807568.html