[leetcode 258]Add Digits

Question：

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?

Code:

方法一：最直观的解法就是去拆解各个位置上的数字，不停求和拆解，直到满足条件 虽然不满足条件，但是还是写出来。
方法二：逐位相加直到小于10
方法三：找规律

1 2 3 4 5 6 7 8 9
10 11 12 13 14 15 16 17 18
19 20 21 22 23 24 25 26 27

每9个一循环, 然而 9除9取模为0，不符合实际
转化为 (n-1)%9 + 1 ;

// 方法一
public class Solution {
    public int addDigits(int num) {
        if(num > 10){
            num = getSum(num);
        }
        return num ;
    }

    /* 求得各个位置的数，并求和 */
    public int getSum(int num){
        int sum = 0 ;
        for(char c : String.valueOf(num).toCharArray()){
            sum += c - '0' ;
        }    
        return sum ; 
    }    
}

// 方法二
public class Solution {
    public int addDigits(int num) {
        while(num > 9){
            num = num/10 + num%10 ;
        }    
        return num ;
    }

}

// 方法三
public class Solution {
    public int addDigits(int num) {
        return (num-1)%9 + 1 ;
    }

}

python

class Solution(object):
    def addDigits(self, num):
        """
        :type num: int
        :rtype: int
        """
        if (num==0):
            return num ;
        return (num-1)%9+1;

JS

/**
 * @param {number} num
 * @return {number}
 */
var addDigits = function(num) {
    return (num-1)%9 +1;
};