【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 - ) / );

    }

};

巴特西

【LeetCode】258. Add Digits (2 solutions)

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