4

I have a program like this

public class no_of_letters_count {
static int i;
 public static void main(String[] args) 
 {

  String sMessage="hello how r u";
  String saMessage[] = sMessage.split("");
  List sList = Arrays.asList(saMessage);                  
  Collections.sort(sList);
  Iterator i=sList.iterator();
  while (i.hasNext())
  {
   System.out.println((String)i.next());
  }
 }
}

//Now i want to count the number of occurance of each characters.

Like

Count of h=2

Count of e=1

and so on

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3
  • Can you reformat your post so it looks better? Commented Feb 1, 2010 at 12:31
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4 Answers 4

Reset to default
4

Iterate over sList and put each char in a HashMap. If it doesn't exists start with count 1, otherwise increment the count.

EDIT : Cannot help posting some code.

First of all, use gnerics. 

List<Character> sList = Arrays.asList(saMessage.toCharArray());

Then use the following map :

Map<Character, Integer> cmap = new HashMap<Character, Integer>();
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1 Comment

I was going to post it, but I won't now :)
1

Using commons-collections and commons-lang

List<Character> chars = Arrays.asList(
     ArrayUtils.toObject("asdfasdas".toCharArray()));
Bag bag = new HashBag(chars);

System.out.println(bag.getCount('a'));
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4 Comments

do you really need Appache-Commons to count the letters? its just a homework
why not? you just add 2 libraries and write 2 lines. If it were my homework, I'd do it this way. On the other hand - yes, it's not the point
In most homework assignments it's stipulated that you must implement the core algorithm yourself rather than import it from a library like this. It's probably okay (but overkill) to use it to iterate over the string, but not to use an off-the-shelf bag/multiset implementation.
it wasn't stipulated in the question :) Anyway, I agree this is missing the point of it being a homework. But since the point is already lost after asking for a solution on SO.. in addition, he will probably have headaches adding the libraries to the classpath, which is again a good learning experience
0

Prints number of occurrences of characters from A-Z and a-z

Note: using the character as array index, not safe if you have Unicode chars :), but still good idea i think!!

String str = "sometext anything";
int[] array=new int[256];

for (int x:array) array[x]=0;

for (char c:str.toCharArray()){
    array[c]++;
}

for (int i=65;i<=90; i++){
    System.out.println("Number of "+(char)i+ "="+array[i]
                + ", number of "+(char)(i+32)+"="+ array[i+32]);
}
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2 Comments

If you're not interested in characters with 0 occurances then why not just use a map where the key is the character... Instead of an array of mostly zeros.
i know, but its just a homework and i wanted to show the simplest way (i guess)
0

Here are some hints:

It seems you are just trying to get the chars in a String. Use String.toCharArray().

Use a Map<Character,Integer> to store the chars and it's occurrences:

foreach char c in sMessage.ToCharArray()
if map.containsKey(c)
  map.put(c, map.get(c) + 1);
else
  map.put(c, 1);

Next sort the map and present the results. I leave you here a snippet to sort the map:

    List<Entry<Character,Integer>> l = new ArrayList<Entry<Character,Integer>>(map.entrySet());

    Collections.sort(l, new Comparator<Entry<Character,Integer>>() {
        public int compare(Entry<Character, Integer> o1, Entry<Character, Integer> o2) {
            return o1.getKey().compareTo(o2.getKey());
        }
    });
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1 Comment

Why not just use a SortedMap to avoid the overhead of the additional sort? Also, you can avoid calling containsKey(c) by calling get(c) and assigning the result to an Integer, and comparing this against null.

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