Add Digits

Add Digits
Difficulty: Easy
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?
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最直接的思路：
num2用来统计num有多少位；
num1用来做除数。
eg.num为345，取百位，则num1得为1000（取余后，/（num1/10）即得百位数3）

    int addDigits(int num) {        int num1,num2;        num1=1,num2=0;        int sum=0;        if(num/10 == 0)            return num;        while(num1 <= num){            num1 *= 10;            num2++;        }        while(num2--){            num %= num1;            num1 /= 10;            sum += num/num1;        }        addDigits(sum);    }

最终结果显示：Last executed input:
2032610959
Status: Time Limit Exceeded

显然该法虽直接，但运算效率太低。
于是我们对这些数做了下计算：
0-9，全部都各自返回本身（0-9）
10 -> 1
11 - > 2
……
19 - > 2
20 - > 2
21 - >　３
……
发现规律，最终结果的区间总落在[0，9]之间。
21%9余3，20%9余2，19%9余1，18%9余0，但实际上18对应结果为9.
所以应该总结为(num-1)%9+1;

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