character checking

if (c=='a' || c=='e' || c=='i' || c=='o' || c=='u') { count++;

When i give the above statement it returns number of vowe开发者_运维百科ls in a word only if the given word is in lowercase (ie: input: friend output 2 vowels). I Want to know even if i give uppercase or mixed it should return number of vowels. How to do it?

One succint, simple way to do this without doing 10 comparisons:

if ("aeiouAEIOU".indexOf(c) != -1) { count++; }

Here there is a complete example - note they turn the string to lower case before checking the vowels.

You can also try with a case insensitive regular expression, example from http://www.shiffman.net/teaching/a2z/regex/:

String regex = "[aeiou]";               
Pattern p = Pattern.compile(regex,Pattern.CASE_INSENSITIVE);   
int vowelcount = 0;
Matcher m = p.matcher(content);         // Create Matcher
while (m.find()) {
  //System.out.print(m.group());
  vowelcount++;
}

If you want to get upper or lowercase a simple approach seems to be the following:

Iterate over all the characters in the word and do the following test:

if (c=='a' || c=='e' || c=='i' || c=='o' || c=='u'
|| c=='A' || c=='E' || c=='I' || c=='O' || c=='U') 
   { count++ };

You can use tests like:

Character.toLowerCase(c) == 'a'

instead.

I tell you one thing, if you give the out put in upper case then in your code accept given out put in lower case.

if string str='FRIEND'

handle str.Tolower(); Please try this I think your problem resoloved.

How about:

if (Arrays.asList('a', 'e', 'i', 'o', 'u').contains(Character.toLowerCase(c))) {
   ...
}

I would also static final the list as well.

Here is a quite simple utility class :

public class Main {
    public static int countChars(String string, Character... characters) {
        return countChars(string, new HashSet<Character>(Arrays.asList(characters)));
    }

    public static int countChars(String string, Set<Character> characters) {
        int count = 0;
        for(int i = 0; i < string.length(); i++){
            if(characters.contains(string.charAt(i))){
                count++;
            }
        }
        return count;
    }

    public static int countCharsIgnoreCase(String string, Character... characters) {
        return countCharsIgnoreCase(string, new HashSet<Character>(Arrays.asList(characters)));
    }

    public static int countCharsIgnoreCase(String string, Set<Character> characters) {
        Set<Character> finalCharacters = new HashSet<Character>();
        for (Character character : characters) {
            finalCharacters.add(Character.toUpperCase(character));
            finalCharacters.add(Character.toLowerCase(character));
        }

        return countChars(string, finalCharacters);
    }
}

code on ideone

Not the most efficient solution, but perhaps the shortest? :)

int vowelCount = (inputString+" ").split("(?i)[aoeuiy]").length - 1

Btw, am I the only one counting 'y' as a wovel? ;)

Didn't see a Set yet :-(

  static Character charvowels[] = { 'A','I','U','E','O','a','i','u','e','o' };
  static Set<Character> vowels = new HashSet<Character>(Arrays.asList(charvowels));
  ...
  public void countVowel(char c) {
    if (vowels.contains(c)) count++;
  }

Using Guava's CharMatcher:

private final CharMatcher vowels = CharMatcher.anyOf("aeiouAEIOU");

...

// somewhere in your method:
if (vowels.matches(c)) {
    ...
}