标签：
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:
Credits:
Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.
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c++ code:
class Solution {
public:
int addDigits(int num) {
return num-9*((num - 1)/9);
}
};标签：
原文地址：http://blog.csdn.net/itismelzp/article/details/51331216
