(leetcode)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?

class Solution {
public:
    int addDigits(int num) {
        int ans = num;
        // while(ans!=2)
        // {
        while(1)
        {
            if(ans < 10) return ans;
            num = ans;
            ans = 0;
            while(num != 0)
            {
                ans += num%10;
                num /= 10;
            }
        }
    }
};

 

O(1)方法

观察如下输入：

输入：1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20

输出：1,2,3,4,5,6,7,8,9,1 , 2 , 3 , 4 , 5 , 6, 7 , 8 , 9 , 1 , 2 

可以看到返回的值就是(num-1)%9+1