Returns a new string version of the current string where all the letters either >= or <= the given char n, are removed.

The given letter may be a-z or A-Z, inclusive. The letters to be removed are case insensitive. If 'more' is false, all letters less than or equal to n will be removed in the returned string. If 'more' is true, all letters greater than or equal to n will be removed in the returned string. If the current string is null, the method should return an empty string. If n is not a letter (and the current string is not null), the method should return an empty string.

Questions: Since the method receives char n, how do I apply for loop to each character? Do I use arrayOfArg and is it correct to use arrayOfArg[i] to store letters more than the letter?

public String filterLetters(char n, boolean more) {
    char [] arrayOfArg = n.toCharArray();
    String myNewString = "";
    String removedChar = "";
    
    if (n == null) {
        return "";
    }
    
    for (int i = 0; i < Character.length; i  ) {
        if (more == false) {
            //all letters less than or equal to n will be removed in the returned string
            if (Character.isLetter(i) == true) continue;
            removedChar = arrayOfArg[i]   ;
            myNewString = arrayOfArg.replace(removedChar,"");
                
        }
        if (more == true) {
            //all letters less than or equal to n will be removed in the returned string
            if (Character.isLetter(i) == true) continue;
            removedChar = arrayOfArg[i] -- ;
            myNewString = arrayOfArg.replace(removedChar,"");
        }
    }
    return myNewString;
        } 

CodePudding user response：

This is a slight modification to the code you requested but the output works properly. I'm sure someone else has a better idea than me. This is just a starting place.

public static String filterLetters(String word, String n, boolean more) {
    String[] alphabet = new String[]{"a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o", "p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z"};
    int charValue = 0;
    //Find Numerical Value of Char n
    for (int i = 0; i < alphabet.length; i  ) {
      if (alphabet[i].equalsIgnoreCase(n)) {
        charValue = i   1;
      }
    }
    if (more) {
      for (int i = charValue; i < alphabet.length; i  ) {
        word = word.replaceAll(alphabet[i],"");
      }
    } else {
      for (int i = charValue; i > 0; i--) {
        word = word.replaceAll(alphabet[i],"");
      }
    }

    
    return word;
  }

CodePudding user response：

Use regex!

public String filterLetters(String str, char n, boolean more) {
    if (str == null || !Character.isLetter(n)) {
        return "";
    }
    String range = more ? n   "-z" : "a-"   n;
    return str.replaceAll("(?i)["   range   "]", "");
}

This works by creating a regular expression as follows:

• if more is true: (?i)[n-z]
• if more is false: (?i)[a-n]

where "n" is the actual char passed in.

(?i) flag turns on case insensitivity.

[a-n] is the range (inclusive) of characters from a to n.
[n-z] is the range (inclusive) of characters from n to z.

Note: The necessary String parameter has been added to the method.