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.
Follow up:
Could you do it without any loop/recursion in O(1) runtime?
解法一、按照定义做，直到结果只剩1位数字。
class Solution {
public:
int addDigits(int num) {
int n = num;
while(n > )
{
int cur = ;
while(n)
{
cur += (n % );
n /= ;
}
n = cur;
}
return n;
}
};
解法二、套公式digital root
class Solution {
public:
int addDigits(int num) {
return num - * floor((num - ) / );
}
};