I can also use a Stringbuffer to store the char

lots of ways, here's one that's easy to understand -

import java.util.*;

public class Oi
{
   public static void main(String[] args)
   {
       List<Character> l = new ArrayList<Character>();
       l.add('c');
       l.add('l');
       l.add('h');
       l.add('l');
       
       if(!l.isEmpty())
       {
           int maxOccurrences = 0;
           char mode = '0';
       
           Map<Character, Integer> m = new HashMap<Character, Integer>();
       
           for(char c : l)
           {
               m.put(c, m.get(c) == null ? 1 : m.get(c) + 1);
           }
       
           for(char c : m.keySet())
           {
               if(m.get(c) > maxOccurrences)
               {
                   maxOccurrences = m.get(c);
                   mode = c;
               }
               System.out.println(c + ":" + m.get(c));
           }
           System.out.println("Most frequent was " + mode +
                             " with " + maxOccurrences + " occurrences.");
       }
   }
}

There are several  of ways of tackling that. The fastest  would be to
have a an int[ 64*1024] where you index by char and count up how often
each char occurs.  Then you scan the array looking for the biggest
number.

That requires a fairly huge array most of which is empty.

Another way would be to create a HashMap where you look up by char to
get the count. You then only have entries for chars that occur.  The
lookup is slower.

Another way would be to collect all the characters to be tested into
an array and sort them. You can then count how many of each char
appears and keep track of the best so far, replacing it if you find a
better one.

see http://mindprod.com/jgloss/biggest.html
for how to find the biggest in an array.