How do I count the number of occurrences of a char in a String?

Question

I have the string

a.b.c.d

I want to count the occurrences of '.' in an idiomatic way, preferably a one-liner.

(Previously I had expressed this constraint as "without a loop", in case you're wondering why everyone's trying to answer without using a loop).

Answer 1

Seems a shame to allocate an array just to count the occurrences of a character. Why not write the loop?

Answer 2

Sooner or later, something has to loop. It's far simpler for you to write the (very simple) loop than to use something like split which is much more powerful than you need.

By all means encapsulate the loop in a separate method, e.g.

public static int countOccurrences(String haystack, char needle)
{
    int count = 0;
    for (int i=0; i < haystack.length(); i++)
    {
        if (haystack.charAt(i) == needle)
        {
             count++;
        }
    }
    return count;
}

Then you don't need have the loop in your main code - but the loop has to be there somewhere.

Answer 3

here is a solution without a loop:

public static int countOccurrences(String haystack, char needle, int i){
    return ((i=haystack.indexOf(needle, i)) == -1)?0:1+countOccurrences(haystack, needle, i+1);}


System.out.println("num of dots is "+countOccurrences("a.b.c.d",'.',0));

well, there is a loop, but it is invisible :-)

-- Yonatan

Answer 4

String s = "a.b.c.d";
int charCount = s.length() - s.replaceAll("\\.", "").length();

ReplaceAll(".") would replace all characters.

PhiLho's solution uses ReplaceAll("[^.]",""), which does not need to be escaped, since [.] represents the character 'dot', not 'any character'.

Answer 5

I had an idea similar to Mladen, but the opposite...

String s = "a.b.c.d";
int charCount = s.replaceAll("[^.]", "").length();
println(charCount);

Answer 6

Okay, inspired my Monatan's solution, here's one which is purely recursive - the only library methods used are length() and charAt(), neither of which do any looping:

public static int countOccurrences(String haystack, char needle)
{
    return countOccurrences(haystack, needle, 0);
}

private static int countOccurrences(String haystack, char needle, int index)
{
    if (index >= haystack.length())
    {
        return 0;
    }

    int contribution = haystack.charAt(index) == needle ? 1 : 0;
    return contribution + countOccurrences(haystack, needle, index+1);
}

Whether recursion counts as looping depends on which exact definition you use, but it's probably as close as you'll get.

I don't know whether most JVMs do tail-recursion these days... if not you'll get the eponymous stack overflow for suitably long strings, of course.

Answer 7

While methods can hide it, there is no way to count without a loop (or recursion). You want to use a char[] for performance reasons though.

public static int count( final String s, final char c ) {
  final char[] chars = s.toCharArray();
  int count = 0;
  for(int i=0; i<chars.length; i++) {
    if (chars[i] == c) {
      count++;
    }
  }
  return count;
}

Using replaceAll (that is RE) does not sound like the best way to go.

Answer 8

Inspired by Jon Skeet, a non-loop version that wont blow your stack. Also useful starting point if you want to use the fork-join framework.

public static int countOccurrences(CharSequeunce haystack, char needle) {
    return countOccurrences(haystack, needle, 0, haystack.length);
}

// Alternatively String.substring/subsequence is relatively efficient
//   on most, but not all, Java library implementations.
private static int countOccurrences(
    CharSequence haystack, char needle, int start, int end
) {
    if (start == end) {
        return 0;
    } else if (start+1 == end) {
        return haystack.charAt(start) == needle ? 1 : 0;
    } else {
        int mid = (end+start)>>>1; // Watch for integer overflow...
        return
            countOccurrences(haystack, needle, start, mid) +
            countOccurrences(haystack, needle, mid, end);
    }
}

(Disclaimer: Not tested, not compiled, not sensible.)

Perhaps the best (single-threaded, no surrogate-pair support) way to write it:

public static int countOccurrences(CharSequeunce haystack, char needle) {
    int count = 0;
    for (char c : haystack.toCharArray()) {
        if (c == needle) {
           ++count;
        }
    }
    return count;
}

Answer 9

Somewhere in the code, something has to loop. The only way around this is a complete unrolling of the loop:

int numDots = 0;
if (s.charAt(0) == '.') {
    numDots++;
}

if (s.charAt(1) == '.') {
    numDots++;
}


if (s.charAt(2) == '.') {
    numDots++;
}

...etc, but then you're the one doing the loop, manually, in the source editor - instead of the computer that will run it. See the pseudocode:

create a project
position = 0
while (not end of string) {
    write check for character at position "position" (see above)
}
write code to output variable "numDots"
compile program
hand in homework
do not think of the loop that your "if"s may have been optimized and compiled to

Answer 10

Here is a slightly different style recursion solution:

public static int countOccurrences(String haystack, char needle)
{
    return countOccurrences(haystack, needle, 0);
}

private static int countOccurrences(String haystack, char needle, int accumulator)
{
    if (haystack.length() == 0) return accumulator;
    return countOccurrences(haystack.substring(1), needle, haystack.charAt(0) == needle ? accumulator + 1 : accumulator);
}

Answer 11

omg java api is from the past. when will we have this very basic feature on the most popular programming language in the world? millions of classes, billions of methods, but there is no a simple plain way to count substring ocurrences.

Answer 12

A shorter example is

String text = "a.b.c.d";
int count = text.split("\\.",-1).length-1;

Answer 13

My 'idiomatic one-liner' for this is:

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

Why write it yourself when it's already in commons lang?

Answer 14

StringTokenizer stOR = new StringTokenizer(someExpression, "||"); int orCount = stOR.countTokens()-1;