Find occurrences of characters in a Java String [duplicate]

Answer 1

      String searchFor = "a";
      String base = "aaaab";
      int count=0;
      int index =base.indexOf(searchFor);

      while(index!=-1){
          ++count;
          index = base.indexOf(searchFor, index+searchFor.length());
      }

      System.out.println(count);

Answer 2

You could use StringUtils class provided by apache commons. StringUtils.countMatches(String originalString, String subCharacterSequesnce)

Answer 3

public static void main(String[] args) {

    Map<Character, Integer> data = new HashMap<Character, Integer>();

    String s = "aaaab";

    char[] chars = s.toCharArray();
    for (char a : chars) {

        if (data.containsKey(a)) {
            int value = data.get(a);
            data.put(a, value + 1);
        } else {
            data.put(a, 1);
        }

    }
    Iterator it = data.entrySet().iterator();
    while (it.hasNext()) {
        Map.Entry pairs = (Map.Entry) it.next();
        System.out.println(pairs.getKey() + " = " + pairs.getValue());
    }
}

Answer 4

Here is a really short solution without any extra libraries:

String input = "aaaab";

int i = -1, count = 0;
while( (i = input.indexOf( 'a', i + 1 ) ) != -1 ) count++;

System.out.println( count );

Answer 5

A simple loop over the characters would do it.

public int countChars(char c, String s) {
  int result = 0;
  for (int i = 0, n = s.length(); i < n; i++) {
    if (s.charAt(i) == c) {
      result++;
    }
  }
  return result;
}

Answer 6

You can simply use this:

String a = "i am here as junior java programmer";
Set temp = new HashSet();
char[] chararray=a.toCharArray();
Set temp=new HashSet();
for(int i=0;i<chararray.length;i++)
{
    int  count=0;
    for (int j=0;j<chararray.length;j++) {
        if (chararray[i]==chararray[j]) {
            count++;
        }            
    }
    if (temp.add(chararray[i])!=false)
        System.out.println("Character "+chararray[i]+" occur "+count);

}