258.整数各位相加，最后得到仅一位的数字 Add Digits

Given an array of integers, every element appears twice except for one. Find that single one.

Note:
Your algorithm should have a linear runtime complexity. Could you implement it without using extra memory?

有一个非负整数num，重复这样的操作：对该数字的各位数字求和，对这个和的各位数字再求和……直到最后得到一个仅1位的数字（即小于10的数字）。

例如：num=38，3+8=11,1+1=2。因为2小于10，因此返回2。

/*
    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? 
 */
using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
namespace Solution {
    public class Solution {
        public int AddDigits(int num) {
            char[] nums = num.ToString().ToCharArray();
            if (num / 10 >= 1) {
                int sum = 0;
                foreach (char c in nums) {
                    sum += (Convert.ToInt32(c) - 48);
                }
                return AddDigits(sum);
            } else {
                return num;
            }
        }
        //one line solution
        public int AddDigits(int num) {
            return (num - 1) % 9 + 1;
        }
    }
    class Program {
        static void Main(string[] args) {
            var s = new Solution();
            var res = s.AddDigits(38);
            Console.WriteLine(res);
        }
    }
}

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