How do I count the number of occurrences of a char

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

Okay, inspired by Yonatan'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 2

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 3

How about this. It doesn't use regexp underneath so should be faster than some of the other solutions and won't use a loop.

int count = line.length() - line.replace(".", "").length();

Answer 4

With java-8 you could also use streams to achieve this. Obviously there is an iteration behind the scenes, but you don't have to write it explicitly!

public static long countOccurences(String s, char c){
    return s.chars().filter(ch -> ch == c).count();
}

countOccurences("a.b.c.d", '.'); //3
countOccurences("hello world", 'l'); //3

Answer 5

Also possible to use reduce in Java 8 to solve this problem:

int res = "abdsd3$asda$asasdd$sadas".chars().reduce(0, (a, c) -> a + (c == '$' ? 1 : 0));
System.out.println(res);

Output:

3

Answer 6

In case you're using Spring framework, you might also use "StringUtils" class. The method would be "countOccurrencesOf".