leetcode_258_Add Digit(easy)(C++)

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?

思路一

逐位相加得到个位数肯定是在1-9之间（n = 0时为0）

class Solution{

public:

int addDigits(int num){

while(num > 9)

{num = num/10 + num % 10;}

return num;

}

}

思路二：

class Solution {
public:
    int addDigits(int num) {
       return 1 + (num - 1)%9;
    }
};