【LeetCode】258. Add Digits (2 solutions)

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 - ) / );
    }
};