[leetcode]258. Add Digits

Given a non-negative integer num, repeatedly add all its digits until the result has only one digit.

Example:

Input: 38Output: 2 
Explanation: 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?

分析：

各位数字相加，直到只有一位即可。若n/10不为0，则一直将各位相加即可。

解法一：

class Solution {
public:
    int addDigits(int num) {
        while(num/10)
        {
          num = num/10 + num%10;  
        }
        return num;
    }
};

如果不用循环，那可能就是有相关规律，计算发现10~18的结果是1~9，所以9个数字一循环。对于小于9的数直接除以9求余数即可；对于大于等于9的数，可以用（n-1）%9+1来解决。

解法2：

class Solution {
public:
    int addDigits(int num) {
        if(num==0)
            return 0;
        else
            return 1+(num-1)%9;
    }
};