【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1、按照定義作，直到結果只剩1位數字。

class Solution {
public:
    int addDigits(int num) {
        int n = num;
        while(n > 9)
        {
            int cur = 0;
            while(n)
            {
                cur += (n % 10);
                n /= 10;
            }
            n = cur;
        }
        return n;
    }
};

 

解法2、套公式digital root

class Solution {
public:
    int addDigits(int num) {
        return num - 9 * floor((num - 1) / 9);
    }
};