leetcode Add Digits

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.

方法一：采用循环：

int addDigits(int num) {    int sum;while(num/10>0){sum=0;while(num>0){sum+=num%10;num/=10;}num=sum;}return num;}

方法二：先来观察1到20的所有的最终返回值：

1------1

2------2

3------3

.

.

.8-----8

9------9

10----1

11-----2

.

.

.

18----9

19----1

20----2

由此可以发现这是以9为循环，1-9，然后是10-18，然后是19-27，以此类推，某个数字n对应返回值(n-1)%9+1:

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