Counting the number of occurences of characters in a string [duplicate]

Answer 1

public class StringTest{
public static void main(String[] args){

    String s ="aaabbbbccccccdd";
    String result="";
    StringBuilder sb = new StringBuilder(s);


    while(sb.length() != 0){
        int count = 0;
        char test = sb.charAt(0);
        while(sb.indexOf(test+"") != -1){
            sb.deleteCharAt(sb.indexOf(test+""));
            count++;
        }
        //System.out.println(test+" is repeated "+count+" number of times");
        result=result+test+count;
    }
    System.out.println(result);         
}
}

Answer 2

I think what you are looking for is this:

public class Ques2 {

/**
 * @param args the command line arguments
 */
public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    String input = br.readLine().toLowerCase();
    StringBuilder result = new StringBuilder();
    char currentCharacter;
    int count;

    for (int i = 0; i < input.length(); i++) {
        currentCharacter = input.charAt(i);
        count = 1;
        while (i < input.length() - 1 && input.charAt(i + 1) == currentCharacter) {
            count++;
            i++;
        }
        result.append(currentCharacter);
        result.append(count);
    }

    System.out.println("" + result);
}

}

Answer 3

I don't want to give out the full code. So I want to give you the challenge and have fun with it. I encourage you to make the code simpler and with only 1 loop.

Basically, my idea is to pair up the characters comparison, side by side. For example, compare char 1 with char 2, char 2 with char 3, and so on. When char N not the same with char (N+1) then reset the character count. You can do this in one loop only! While processing this, form a new string. Don't use the same string as your input. That's confusing.

Remember, making things simple counts. Life for developers is hard enough looking at complex code.

Have fun!

Tommy "I should be a Teacher" Kwee

Answer 4

if this is a real program and not a study project, then look at using the Apache Commons StringUtils class - particularly the countMatches method.

If it is a study project then keep at it and learn from your exploring :)

Answer 5

This is the problem:

while(str.charAt(i)==ch)

That will keep going until it falls off the end... when i is the same as the length of the string, it will be asking for a character beyond the end of the string. You probably want:

while (i < str.length() && str.charAt(i) == ch)

You also need to set count to 0 at the start of each iteration of the bigger loop - the count resets, after all - and change

count = count + i;

to either:

count++;

... or get rid of count or i. They're always going to have the same value, after all. Personally I'd just use one variable, declared and initialized inside the loop. That's a general style point, in fact - it's cleaner to declare local variables when they're needed, rather than declaring them all at the top of the method.

However, then your program will loop forever, as this doesn't do anything useful:

str.substring(count);

Strings are immutable in Java - substring returns a new string. I think you want:

str = str.substring(count);

Note that this will still output "a2b2a2" for "aabbaa". Is that okay?

Answer 6

You should be able to utilize the StringUtils class and the countMatches() method.

public static int countMatches(String str, String sub)

Counts how many times the substring appears in the larger String.

Try the following:

int count = StringUtils.countMatches("a.b.c.d", ".");