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, thePRocess is like: 3 + 8 = 11, 1 + 1 = 2. Since 2 has only one digit, return it.
思路：
除和余
packageothers;publicclassAddDigits {publicintaddDigits(intnum) {if(num / 10 == 0)returnnum;intres = 0;while(num > 0) {
res+= num % 10;
num= num / 10;
}returnaddDigits(res);
}publicstaticvoidmain(String[] args) {//TODO Auto-generated method stub}
}