Brooke

一、題目

You are given a string s formed by digits and '#'. We want to map s to English lowercase characters as follows:

• Characters ('a' to 'i') are represented by ('1' to '9') respectively.
• Characters ('j' to 'z') are represented by ('10#' to '26#') respectively.

Return the string formed after mapping.

The test cases are generated so that a unique mapping will always exist.

 

Example 1:

Input: s = "10#11#12" Output: "jkab" Explanation: "j" -> "10#" , "k" -> "11#" , "a" -> "1" , "b" -> "2".

Example 2:

Input: s = "1326#" Output: "acz"

 

Constraints:

• 1 <= s.length <= 1000
• s consists of digits and the '#' letter.
• s will be a valid string such that mapping is always possible.

 

二、程式作法

public class Solution
{
    public string FreqAlphabets(string s)
    {
        string res = "";
        string[] letters = s.Split("#");

        for (int i = 0; i < letters.Length; i++)
        {
            if (i == letters.Length - 1)
            {
                char[] c = letters[i].ToCharArray();
                for (int j = 0; j < c.Length; j++)
                    res += ToConvert(c[j].ToString());
            }
            else
            {
                char[] c = letters[i].Substring(0, letters[i].Length - 2).ToCharArray();
                for (int j = 0; j < c.Length; j++)
                    res += ToConvert(c[j].ToString());
                res += ToConvert(letters[i].Substring(letters[i].Length - 2));
            }
        }

        return res;
    }

    public string ToConvert(string a)
    {
        return ((char)(Convert.ToInt32(a) + 96)).ToString();
    }
}

 

public class Solution
{
    public string FreqAlphabets(string s)
    {
        string[] ss = s.Split("#");
        for (int i = 0; i < ss.Length; i++)
        {
            char[] cs = ss[i].ToCharArray();
            for (int j = 0; j < cs.Length; j++)
                if (j < cs.Length - 2 || i == ss.Length - 1)
                    cs[j] = (char)(cs[j] + 48);
                else
                {
                    cs[j] = (char)(int.Parse(string.Concat(cs[j], cs[j + 1])) + 96);
                    cs = cs.Where((val, idx) => idx < cs.Length - 1).ToArray();
                }
            ss[i] = new string(cs);
        }

        return string.Concat(ss);
    }
}

 

三、思路筆記

針對 # 字號做切割，除了最後一段之外，每一段長度至少會有 j ~ z 其中一字母。

做數字與字母之間的轉換時，可利用字母跟 ASCII code 的關係。