Optimized the answer of Sarvar Nishonboev, seems like a simplest solution without unnecessary complexity related to the re-creating of string at each iteration:
public static class Converter
{
private static readonly Dictionary<char, string> ConvertedLetters = new Dictionary<char, string>
{
{'а', "a"},
{'б', "b"},
{'в', "v"},
{'г', "g"},
{'д', "d"},
{'е', "e"},
{'ё', "yo"},
{'ж', "zh"},
{'з', "z"},
{'и', "i"},
{'й', "j"},
{'к', "k"},
{'л', "l"},
{'м', "m"},
{'н', "n"},
{'о', "o"},
{'п', "p"},
{'р', "r"},
{'с', "s"},
{'т', "t"},
{'у', "u"},
{'ф', "f"},
{'х', "h"},
{'ц', "c"},
{'ч', "ch"},
{'ш', "sh"},
{'щ', "sch"},
{'ъ', "j"},
{'ы', "i"},
{'ь', "j"},
{'э', "e"},
{'ю', "yu"},
{'я', "ya"},
{'А', "A"},
{'Б', "B"},
{'В', "V"},
{'Г', "G"},
{'Д', "D"},
{'Е', "E"},
{'Ё', "Yo"},
{'Ж', "Zh"},
{'З', "Z"},
{'И', "I"},
{'Й', "J"},
{'К', "K"},
{'Л', "L"},
{'М', "M"},
{'Н', "N"},
{'О', "O"},
{'П', "P"},
{'Р', "R"},
{'С', "S"},
{'Т', "T"},
{'У', "U"},
{'Ф', "F"},
{'Х', "H"},
{'Ц', "C"},
{'Ч', "Ch"},
{'Ш', "Sh"},
{'Щ', "Sch"},
{'Ъ', "J"},
{'Ы', "I"},
{'Ь', "J"},
{'Э', "E"},
{'Ю', "Yu"},
{'Я', "Ya"}
};
public static string ConvertToLatin(string source)
{
var result = new StringBuilder();
foreach (var letter in source)
{
result.Append(ConvertedLetters[letter]);
}
return result.ToString();
}
}
Use it like this:
Converter.ConvertToLatin("Проверочный текст");