[C#][LeetCode][Easy] 258. Add Digits

心得：

題目要求將每一位數相加，直到剩下個位數。

問題：

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.

答案：

1. For + 遞迴

   public class Solution {
       public int AddDigits(int num) {
           string str = num.ToString();
           if (str.Length < 2)
           {
               return int.Parse(str);
           }
           else
           {
               int tmp = 0;
               for (int i = 0; i < str.Length; i++)
               {
                   tmp += int.Parse(str[i].ToString());
               }
       
               return AddDigits(tmp);
           }
       }
   }

2. LinQ + 遞迴

   public class Solution {
       public int AddDigits(int num) {
           string str = num.ToString();
           if (str.Length < 2)
           {
               return int.Parse(str);
           }
           else
           {
               return AddDigits(str.Sum(x => int.Parse(x.ToString())));
           }
       }
   }

3. Normal

   public class Solution {
       public int AddDigits(int num) {
           string str = num.ToString();
           while (str.Length > 1)
           {
               int tmp = 0;
               for (int i = 0; i < str.Length; i++)
               {
                   tmp += int.Parse(str[i].ToString());
               }
   
               str = tmp.ToString();
           }
   
           return int.Parse(str);
       }
   }