How do I count the number of occurrences of a char

2018-12-31 01:46发布

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).

标签: java string
30条回答
情到深处是孤独
2楼-- · 2018-12-31 02:47

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

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不再属于我。
3楼-- · 2018-12-31 02:47

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
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大哥的爱人
4楼-- · 2018-12-31 02:48

Not sure about the efficiency of this, but it's the shortest code I could write without bringing in 3rd party libs:

public static int numberOf(String target, String content)
{
    return (content.split(target).length - 1);
}
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旧时光的记忆
5楼-- · 2018-12-31 02:49

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 use to be relatively efficient
//   on most Java library implementations, but isn't any more [2013].
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(String haystack, char needle) {
    int count = 0;
    for (char c : haystack.toCharArray()) {
        if (c == needle) {
           ++count;
        }
    }
    return count;
}
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余生请多指教
6楼-- · 2018-12-31 02:49

The following source code will give you no.of occurrences of a given string in a word entered by user :-

import java.util.Scanner;

public class CountingOccurences {

    public static void main(String[] args) {

        Scanner inp= new Scanner(System.in);
        String str;
        char ch;
        int count=0;

        System.out.println("Enter the string:");
        str=inp.nextLine();

        while(str.length()>0)
        {
            ch=str.charAt(0);
            int i=0;

            while(str.charAt(i)==ch)
            {
                count =count+i;
                i++;
            }

            str.substring(count);
            System.out.println(ch);
            System.out.println(count);
        }

    }
}
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余生无你
7楼-- · 2018-12-31 02:50

Well, with a quite similar task I stumbled upon this Thread. I did not see any programming language restriction and since groovy runs on a java vm: Here is how I was able to solve my Problem using Groovy.

"a.b.c.".count(".")

done.

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