LeetCode OJ 之 Add Digits （数字相加）

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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?

思路：

1、常规解法：循环或者递归

2、数字根问题，参考维基百科的证明和结论：https://en.wikipedia.org/wiki/Digital_root#Congruence_formula.

For base b (decimal case b = 10), the digit root of an integer is:

dr(n) = 0 if n == 0
dr(n) = (b-1) if n != 0 and n % (b-1) == 0
dr(n) = n mod (b-1) if n % (b-1) != 0

dr(n) = 1 + (n - 1) % 9

其实通过观察前20个数可以发现结果1，2，3……9是周期性出现的，可以得到关系if(n%9 == 0) return 9. if(n%9!=0) return n%9。

代码1：

class Solution {public:    int addDigits(int num)     {        if(num <= 0)            return 0;        while(num >= 10)        {            int tmp = 0;            while(num >= 10)            {                tmp += num % 10;                num = num / 10;            }            tmp += num;            num = tmp;        }        return num;    }};

代码2：

class Solution {public:    int addDigits(int num)     {        if(num <= 0)            return 0;        if(num % 9 != 0)            return num % 9;        else             return 9;    }};

代码3：

class Solution {public:    int addDigits(int num)     {        if(num <= 0)            return 0;        return 1 + (num - 1 ) % 9;    }};

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