How do I convert a number to a letter in Java

Question

Is there a nicer way of converting a number to its alphabetic equivalent than this   private String getCharForNumber int i        char   alphabet    ABCDEFGHIJKLMNOPQRSTUVWXYZ  toCharArray        if  i  gt  25            return null            return Character toString alphabet i        Maybe something than can deal with numbers greater than 26 more elegantly too

User · Answer

if you define a A as 0  char res  if  i gt 25    i lt 0       res   null        res    i    65   return res    65 for captitals  97 for non captitals

User · Answer

Rather than giving an error or some sentinel value  e g       for inputs outside of 0-25  I sometimes find it useful to have a well-defined string for all integers  I like to use the following      0 - gt     A    1 - gt     B    2 - gt     C        25 - gt     Z   26 - gt    AA   27 - gt    AB   28 - gt    AC       701 - gt    ZZ  702 - gt   AAA        This can be extended to negatives as well     -1 - gt    -A   -2 - gt    -B   -3 - gt    -C       -26 - gt    -Z  -27 - gt   -AA        Java Code   public static String toAlphabetic int i        if  i lt 0             return  -  toAlphabetic -i-1              int quot   i 26      int rem   i 26      char letter    char   int  A    rem       if  quot    0             return    letter        else           return toAlphabetic quot-1    letter            Python code  including the ability to use alphanumeric  base 36  or case-sensitive  base 62  alphabets   def to alphabetic i base 26       if base  lt  0 or 62  lt  base          raise ValueError  Invalid base        if i  lt  0          return  -  to alphabetic -i-1       quot   int i  base     rem   i base     if rem  lt  26          letter   chr  ord  A     rem      elif rem  lt  36          letter   str  rem-26      else          letter   chr  ord  a     rem - 36      if quot    0          return letter     else          return to alphabetic quot-1 base    letter

User · Answer

I would return a character char instead of a string   public static char getChar int i        return i lt 0    i gt 25          char   A    i       Note  when the character decoder doesn t recognise a character it returns    I would use  A  or  a  instead of looking up ASCII codes

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public static void main String   args          int rem n 702 quo       String s           while n gt 0                  rem  n-1  26           quo  n-1  26           s  char  rem 97  s           if quo  1                         s  char  97  s              break                      else          n  n-1  26              System out print s            We can also write the code like the below one  There is no much difference but it may help to understand the concept for some people    public static void main String   args          int rem n 52 quo       String s           while n gt 0                  rem n 26           quo n 26           if rem  0               rem 26           s  char  rem 96  s           if  quo  1    quo  0   amp  amp  n gt 26                          n n 26              s  char  n 96  s              break                      else              n n 26-1               System out print s

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Another approach starting from 0 and returning a String  public static String getCharForNumber int i        return i  lt  0    i  gt  25         String valueOf  char    A    i

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Personally  I prefer  return  ABCDEFGHIJKLMNOPQRSTUVWXYZ  substring i  i 1     which shares the backing char     Alternately  I think the next-most-readable approach is  return Character toString  char   i    A       which doesn t depend on remembering ASCII tables   It doesn t do validation  but if you want to  I d prefer to write  char c    char   i    A    return Character isUpperCase c    Character toString c    null    just to make it obvious that you re checking that it s an alphabetic character

User · Answer

You can try like this   private String getCharForNumber int i        CharSequence css    ABCDEFGHIJKLMNOPQRSTUVWXYZ       if  i  gt  25            return null            return css charAt i

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You can convert the input to base 26  Hexavigesimal  and convert each  digit  back to base 10 individually and apply the ASCII mapping  Since A is mapped to 0  you will get results A  B  C      Y  Z  BA  BB  BC    etc  which may or may not be desirable depending on your requirements for input values   26  since it may be natural to think AA comes after Z   public static String getCharForNumber int i           return null for bad input     if i  lt  0           return null                convert to base 26     String s   Integer toString i  26        char   characters   s toCharArray         String result           for char c   characters              convert the base 26 character back to a base 10 integer         int x   Integer parseInt Character valueOf c  toString    26              append the ASCII value to the result         result    String valueOf  char  x    A                          return result

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for int i 0 i lt ar length   i                  char ch   ar charAt i                       System out println  char  ch 16

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This isn t exactly an answer  but some useful related code I made  When you run it and enter any character in the command line it returns getNumericValue char      which doesn t seem to be the same as the ASCII table so be aware  Anyways  not a direct answer to your question but hopefully helpful   import java lang    import java util       charVal java                infinite loops  whenever you type in a character gives you value of    getNumericValue char       ctrl c to exit    public class charVal      public static void main String   args         Scanner inPut   new Scanner System in        for             char c   inPut next   charAt 0          System out printf   n  s    d  n   c  Character getNumericValue c

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Another variant   private String getCharForNumber int i        if  i  gt  25    i  lt  0            return null            return new Character  char   i   65   toString

User · Answer

Just make use of the ASCII representation   private String getCharForNumber int i        return i  gt  0  amp  amp  i  lt  27   String valueOf  char  i   64     null      Note  This assumes that i is between 1 and 26 inclusive    You ll have to change the condition to i  gt  -1  amp  amp  i  lt  26 and the increment to 65 if you want i to be zero-based   Here is the full ASCII table  in case you need to refer to         Edit   As some folks suggested here  it s much more readable to directly use the character  A  instead of its ASCII code   private String getCharForNumber int i        return i  gt  0  amp  amp  i  lt  27   String valueOf  char  i    A  - 1     null

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public static String abcBase36 int i        char   ALPHABET    0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZ  toCharArray        int quot   i   36      int rem   i   36       char letter   ALPHABET rem       if  quot    0            return      letter        else           return abcBase36 quot - 1    letter

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public static string IntToLetters int value    string result   string Empty  while  --value  gt   0        result    char   A    value   26     result      value    26    return result      To meet the requirement of A being 1 instead of 0  I ve added -- to the while loop condition  and removed the value-- from the end of the loop  if anyone wants this to be 0 for their own purposes  you can reverse the changes  or simply add value    at the beginning of the entire method

User · Answer

Getting the alphabetical value from an int can be simply done with    char        i

[java] How do I convert a number to a letter in Java?

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