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

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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  gt   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

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A much easier solution would be to just the split the string based on the character you re matching it with   For instance   int getOccurences String characters  String string        String   words   string split characters       return words length - 1     This will return 4 in the case of  getOccurences  o    something about a quick brown fox

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What about below recursive algo Which is also linear time     import java lang    import java util     class longestSubstr   public static void main String   args      String s  ABDEFGABEF        int ans calc s       System out println  Max nonrepeating seq    ans       public static int calc String s     s s       int n s length          int max 1        if n  1            return 1        if n  2                    if s charAt 0   s charAt 1   return 1            else return 2                  String s1 s      String a s charAt n-1                s1 s1 replace a                  System out println s      n-2      s substring 0 n-1             max Math max calc s substring 0 n-1    calc s1  1      return max           lt  i gt

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public static int countSubstring String subStr  String str         int count   0      for  int i   0  i  lt  str length    i              if  str substring i  startsWith subStr                 count                        return count

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String   parts   text split       int occurances   parts length - 1     It s a great day at O S G  Dallas         -- Famous Last Words   Well  it s a case of knowing your Java  especially your basic foundational understanding of the collection classes already available in Java  If you look throughout the entire posting here  there is just about everything short of Stephen Hawking s explanation of the Origin of the Universe  Darwin s paperback on Evolution and Gene Roddenberry s Star Trek cast selection as to why they went with William Shatner short of how to do this quickly and easily         need I say anymore

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public static void getCharacter String str            int count    new int 256            for int i 0 i lt str length    i                   count str charAt i                         System out println  The ascii values are    Arrays toString count               Now display wht character is repeated how many times          for  int i   0  i  lt  count length  i                  if  count i   gt  0                 System out println  Number of      char  i          count i

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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   Spring Framework s oneliner for this is   int occurance   StringUtils countOccurrencesOf  a b c d

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String s    a b c d   long result   s chars   filter ch - gt  ch         count

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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  lt  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

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public class OccurencesInString   public static void main String   args    String str    NARENDRA AMILINENI   HashMap occur   new HashMap    int count  0  String key   null  for int i 0 i lt str length  -1 i     key   String valueOf str charAt i    if occur containsKey key    count    Integer occur get key   occur put key   count    else  occur put key 1       System out println occur

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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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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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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  gt   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

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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 - gt  ch    c  count       countOccurences  a b c d           3 countOccurences  hello world    l      3

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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 lt chars length  i          if  chars i     c          count                return count      Using replaceAll  that is RE  does not sound like the best way to go

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public static void getCharacter String str            int count    new int 256            for int i 0 i lt str length    i                   count str charAt i                         System out println  The ascii values are    Arrays toString count               Now display wht character is repeated how many times          for  int i   0  i  lt  count length  i                  if  count i   gt  0                 System out println  Number of      char  i          count i

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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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Complete sample   public class CharacterCounter      public static int countOccurrences String find  String string          int count   0      int indexOf   0       while  indexOf  gt  -1              indexOf   string indexOf find  indexOf   1         if  indexOf  gt  -1          count               return count          Call   int occurrences   CharacterCounter countOccurrences  l    Hello World     System out println occurrences      3

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This is what I use to count the occurrences of a string  Hope someone finds it helpful      private long countOccurrences String occurrences  char findChar           return  occurrences chars   filter  x - gt                return x    findChar             count

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import java util Scanner   class apples        public static void main String args                  Scanner bucky   new Scanner System in           String hello   bucky nextLine            int charCount   hello length   - hello replaceAll  e       length            System out println charCount                  COUNTS NUMBER OF  e  CHAR  s within any string input

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I tried to work out your question with a switch statement but I still required a for loop to parse the string   feel free to comment if I can improve the code  public class CharacterCount   public static void main String args          String message  hello how are you       char   array message toCharArray        int a 0      int b 0      int c 0      int d 0      int e 0      int f 0      int g 0      int h 0      int i 0      int space 0      int j 0      int k 0      int l 0      int m 0      int n 0      int o 0      int p 0      int q 0      int r 0      int s 0      int t 0      int u 0      int v 0      int w 0      int x 0      int y 0      int z 0        for char element array                switch element                    case  a           a            break          case  b           b            break          case  c  c            break           case  d  d            break          case  e  e            break          case  f  f            break           case  g  g            break          case  h           h            break          case  i  i            break          case  j  j            break          case  k  k            break          case  l  l            break          case  m  m            break          case  n  m            break          case  o  o            break          case  p  p            break          case  q  q            break          case  r  r            break          case  s  s            break          case  t  t            break          case  u  u            break          case  v  v            break          case  w  w            break          case  x  x            break          case  y  y            break          case  z  z            break          case     space            break          default  break                      System out println  A   a   B    b    C   c   D   d   E   e   F   f   G   g   H   h       System out println  I   i   J   j   K   k   L   l   M   m   N   n   O   o   P   p       System out println  Q   q   R   r   S   s   T   t   U   u   V   v   W   w   X   x   Y   y   Z   z       System out println  SPACE   space

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You can use the split   function in just one line code   int noOccurence string split     -1  length-1

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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  gt  gt  gt 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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My  idiomatic one-liner  solution   int count    a b c d  length   -  a b c d  replace          length      Have no idea why a solution that uses StringUtils is accepted

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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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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

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I see a lot of tricks and such being used  Now I m not against beautiful tricks but personally I like to simply call the methods that are meant to do the work  so I ve created yet another answer   Note that if performance is any issue  use Jon Skeet s answer instead  This one is a bit more generalized and therefore slightly more readable in my opinion  and of course  reusable for strings and patterns    public static int countOccurances char c  String input        return countOccurancesOfPattern Pattern quote Character toString c    input      public static int countOccurances String s  String input        return countOccurancesOfPattern Pattern quote s   input      public static int countOccurancesOfPattern String pattern  String input        Matcher m   Pattern compile pattern  matcher input       int count   0      while  m find              count              return count

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My  idiomatic one-liner  solution   int count    a b c d  length   -  a b c d  replace          length      Have no idea why a solution that uses StringUtils is accepted

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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 lt chars length  i          if  chars i     c          count                return count      Using replaceAll  that is RE  does not sound like the best way to go

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Why are you trying to avoid the loop  I mean you can t count the  numberOf  dots without checking every single character of the string  and if you call any function  somehow it will loop  This is  String replace should do a loop verifying if the string appears so it can replace every single occurrence   If you re trying to reduce resource usage  you won t do it like that because you re creating a new String just for counting the dots   Now if we talk about the recursive  enter code here  method  someone said that it will fail due to an OutOfMemmoryException  I think he forgot StackOverflowException   So my method would be like this  I know it is like the others but  this problem requires  the loop    public static int numberOf String str int c        int res 0      if str  null          return res      for int i 0 i lt str length   i            if c  str charAt i               res        return res

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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  lt  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

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I had an idea similar to Mladen  but the opposite     String s    a b c d   int charCount   s replaceAll             length    println charCount

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I see a lot of tricks and such being used  Now I m not against beautiful tricks but personally I like to simply call the methods that are meant to do the work  so I ve created yet another answer   Note that if performance is any issue  use Jon Skeet s answer instead  This one is a bit more generalized and therefore slightly more readable in my opinion  and of course  reusable for strings and patterns    public static int countOccurances char c  String input        return countOccurancesOfPattern Pattern quote Character toString c    input      public static int countOccurances String s  String input        return countOccurancesOfPattern Pattern quote s   input      public static int countOccurancesOfPattern String pattern  String input        Matcher m   Pattern compile pattern  matcher input       int count   0      while  m find              count              return count

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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

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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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a lambda one-liner w o the need of an external library  Creates a map with the count of each character  Map lt Character Long gt  counts    quot a b c d quot  codePoints   boxed   collect      groupingBy  t - gt   char  int t  counting         gets   a 1  b 1  c 1  d 1    3  the count of a certain character eg      is given over  counts get         I also write a lambda solution out of morbid curiosity to find out how slow my solution is  preferably from the man with the 10-line solution

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

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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  gt  gt  gt 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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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  gt  gt  gt 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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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

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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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Try this method   StringTokenizer stOR   new StringTokenizer someExpression         int orCount   stOR countTokens  -1

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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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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

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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

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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

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Also possible to use reduce in Java 8 to solve this problem   int res    abdsd3 asda asasdd sadas  chars   reduce 0   a  c  - gt  a    c          1   0    System out println res     Output   3

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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 lt chars length  i          if  chars i     c          count                return count      Using replaceAll  that is RE  does not sound like the best way to go

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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  lt  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

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Why not just split on the character and then get the length of the resulting array  array length will always be number of instances   1   Right

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String   parts   text split       int occurances   parts length - 1     It s a great day at O S G  Dallas         -- Famous Last Words   Well  it s a case of knowing your Java  especially your basic foundational understanding of the collection classes already available in Java  If you look throughout the entire posting here  there is just about everything short of Stephen Hawking s explanation of the Origin of the Universe  Darwin s paperback on Evolution and Gene Roddenberry s Star Trek cast selection as to why they went with William Shatner short of how to do this quickly and easily         need I say anymore

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I tried to work out your question with a switch statement but I still required a for loop to parse the string   feel free to comment if I can improve the code  public class CharacterCount   public static void main String args          String message  hello how are you       char   array message toCharArray        int a 0      int b 0      int c 0      int d 0      int e 0      int f 0      int g 0      int h 0      int i 0      int space 0      int j 0      int k 0      int l 0      int m 0      int n 0      int o 0      int p 0      int q 0      int r 0      int s 0      int t 0      int u 0      int v 0      int w 0      int x 0      int y 0      int z 0        for char element array                switch element                    case  a           a            break          case  b           b            break          case  c  c            break           case  d  d            break          case  e  e            break          case  f  f            break           case  g  g            break          case  h           h            break          case  i  i            break          case  j  j            break          case  k  k            break          case  l  l            break          case  m  m            break          case  n  m            break          case  o  o            break          case  p  p            break          case  q  q            break          case  r  r            break          case  s  s            break          case  t  t            break          case  u  u            break          case  v  v            break          case  w  w            break          case  x  x            break          case  y  y            break          case  z  z            break          case     space            break          default  break                      System out println  A   a   B    b    C   c   D   d   E   e   F   f   G   g   H   h       System out println  I   i   J   j   K   k   L   l   M   m   N   n   O   o   P   p       System out println  Q   q   R   r   S   s   T   t   U   u   V   v   W   w   X   x   Y   y   Z   z       System out println  SPACE   space

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Try this code   package com java test   import java util HashMap  import java util Map   public class TestCuntstring        public static void main String   args             String name    Bissssmmayaa           char   ar   new char name length             for  int i   0  i  lt  name length    i                  ar i    name charAt i                     Map lt Character  String gt  map new HashMap lt Character  String gt             for  int i   0  i  lt  ar length  i                  int count 0              for  int j   0  j  lt  ar length  j                      if ar i   ar j                        count                                                map put ar i   count   no of times                      System out println map

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Why not just split on the character and then get the length of the resulting array  array length will always be number of instances   1   Right

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use a lambda function which removes all characters to count the count is the difference of the before-length and after-length  String s    a b c d   int count   s length   - deleteChars apply  s        length        3   find deleteChars here  if you have to count the occurrences of more than one character it can be done in one swoop  eg  for b c and     int count   s length   - deleteChars apply  s   bc     length        5

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public static int countOccurrences String container  String content       int lastIndex  currIndex   0  occurrences   0      while true            lastIndex   container indexOf content  currIndex           if lastIndex    -1                break                    currIndex   lastIndex   content length            occurrences              return occurrences

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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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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  gt   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

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Using Java 8 and a HashMap without any library for counting all the different chars   private static void countChars String string        HashMap lt Integer  Integer gt  hm   new HashMap lt Integer  Integer gt         string chars   forEach letter - gt  hm put letter   hm containsKey letter    hm get letter    0    1        hm forEach  c  i  - gt  System out println   char c intValue            i

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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 - gt  ch    c  count       countOccurences  a b c d           3 countOccurences  hello world    l      3

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I don t like the idea of allocating a new string for this purpose  And as the string already has a char array in the back where it stores it s value  String charAt   is practically free   for int i 0 i lt s length   num   s charAt i     delim 1 0     does the trick  without additional allocations that need collection  in 1 line or less  with only J2SE

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Using Eclipse Collections  int count   Strings asChars  a b c d   count c - gt  c            If you have more than one character to count  you can use a CharBag as follows   CharBag bag   Strings asChars  a b c d   toBag    int count   bag occurrencesOf         Note  I am a committer for Eclipse Collections

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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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public class OccurencesInString   public static void main String   args    String str    NARENDRA AMILINENI   HashMap occur   new HashMap    int count  0  String key   null  for int i 0 i lt str length  -1 i     key   String valueOf str charAt i    if occur containsKey key    count    Integer occur get key   occur put key   count    else  occur put key 1       System out println occur

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Using Eclipse Collections  int count   Strings asChars  a b c d   count c - gt  c            If you have more than one character to count  you can use a CharBag as follows   CharBag bag   Strings asChars  a b c d   toBag    int count   bag occurrencesOf         Note  I am a committer for Eclipse Collections

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A much easier solution would be to just the split the string based on the character you re matching it with   For instance   int getOccurences String characters  String string        String   words   string split characters       return words length - 1     This will return 4 in the case of  getOccurences  o    something about a quick brown fox

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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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int count    line length   - line replace  str       length     str  length

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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  lt  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

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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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public static int countOccurrences String container  String content       int lastIndex  currIndex   0  occurrences   0      while true            lastIndex   container indexOf content  currIndex           if lastIndex    -1                break                    currIndex   lastIndex   content length            occurrences              return occurrences

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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   Spring Framework s oneliner for this is   int occurance   StringUtils countOccurrencesOf  a b c d

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

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What about below recursive algo Which is also linear time     import java lang    import java util     class longestSubstr   public static void main String   args      String s  ABDEFGABEF        int ans calc s       System out println  Max nonrepeating seq    ans       public static int calc String s     s s       int n s length          int max 1        if n  1            return 1        if n  2                    if s charAt 0   s charAt 1   return 1            else return 2                  String s1 s      String a s charAt n-1                s1 s1 replace a                  System out println s      n-2      s substring 0 n-1             max Math max calc s substring 0 n-1    calc s1  1      return max           lt  i gt

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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 lt chars length  i          if  chars i     c          count                return count      Using replaceAll  that is RE  does not sound like the best way to go

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You can use the split   function in just one line code   int noOccurence string split     -1  length-1

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The simplest way to get the answer is as follow   public static void main String   args        String string    a b c d       String   splitArray   string split       -1       System out println  No of   chars is        splitArray length-1

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public static int countSubstring String subStr  String str         int count   0      for  int i   0  i  lt  str length    i              if  str substring i  startsWith subStr                 count                        return count

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If you want to count the no  of same character in a string  SELENIUM  or you want to print the unique characters of the string  SELENIUM    public class Count Characters In String        public static void main String   args            String s    SELENIUM           System out println s           int counter          String g                for  int i 0  i lt s length    i                 if g indexOf s charAt i      - 1              g g s charAt i                                 System out println g                    for  int i 0  i lt g length    i                           System out print                  System out print s charAt i                     counter 0             for  int j 0  j lt s length    j                 if  g charAt i     s charAt j                 counter counter 1                                             System out print counter                                               OUTPUT                           SELENIUM  SELNIUM   S   1 E   2 L   1 E   1 N   1 I   1 U   1

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import java util Scanner   class apples        public static void main String args                  Scanner bucky   new Scanner System in           String hello   bucky nextLine            int charCount   hello length   - hello replaceAll  e       length            System out println charCount                  COUNTS NUMBER OF  e  CHAR  s within any string input

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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

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A shorter example is  String text    a b c d   int count   text split       -1  length-1

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If you want to count the no  of same character in a string  SELENIUM  or you want to print the unique characters of the string  SELENIUM    public class Count Characters In String        public static void main String   args            String s    SELENIUM           System out println s           int counter          String g                for  int i 0  i lt s length    i                 if g indexOf s charAt i      - 1              g g s charAt i                                 System out println g                    for  int i 0  i lt g length    i                           System out print                  System out print s charAt i                     counter 0             for  int j 0  j lt s length    j                 if  g charAt i     s charAt j                 counter counter 1                                             System out print counter                                               OUTPUT                           SELENIUM  SELNIUM   S   1 E   2 L   1 E   1 N   1 I   1 U   1

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The simplest way to get the answer is as follow   public static void main String   args        String string    a b c d       String   splitArray   string split       -1       System out println  No of   chars is        splitArray length-1

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Try this method   StringTokenizer stOR   new StringTokenizer someExpression         int orCount   stOR countTokens  -1

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I had an idea similar to Mladen  but the opposite     String s    a b c d   int charCount   s replaceAll             length    println charCount

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int count    line length   - line replace  str       length     str  length

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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  gt  gt  gt 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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Why are you trying to avoid the loop  I mean you can t count the  numberOf  dots without checking every single character of the string  and if you call any function  somehow it will loop  This is  String replace should do a loop verifying if the string appears so it can replace every single occurrence   If you re trying to reduce resource usage  you won t do it like that because you re creating a new String just for counting the dots   Now if we talk about the recursive  enter code here  method  someone said that it will fail due to an OutOfMemmoryException  I think he forgot StackOverflowException   So my method would be like this  I know it is like the others but  this problem requires  the loop    public static int numberOf String str int c        int res 0      if str  null          return res      for int i 0 i lt str length   i            if c  str charAt i               res        return res

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A shorter example is  String text    a b c d   int count   text split       -1  length-1

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I don t like the idea of allocating a new string for this purpose  And as the string already has a char array in the back where it stores it s value  String charAt   is practically free   for int i 0 i lt s length   num   s charAt i     delim 1 0     does the trick  without additional allocations that need collection  in 1 line or less  with only J2SE

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a lambda one-liner w o the need of an external library  Creates a map with the count of each character  Map lt Character Long gt  counts    quot a b c d quot  codePoints   boxed   collect      groupingBy  t - gt   char  int t  counting         gets   a 1  b 1  c 1  d 1    3  the count of a certain character eg      is given over  counts get         I also write a lambda solution out of morbid curiosity to find out how slow my solution is  preferably from the man with the 10-line solution

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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   gt 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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use a lambda function which removes all characters to count the count is the difference of the before-length and after-length  String s    a b c d   int count   s length   - deleteChars apply  s        length        3   find deleteChars here  if you have to count the occurrences of more than one character it can be done in one swoop  eg  for b c and     int count   s length   - deleteChars apply  s   bc     length        5

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Also possible to use reduce in Java 8 to solve this problem   int res    abdsd3 asda asasdd sadas  chars   reduce 0   a  c  - gt  a    c          1   0    System out println res     Output   3

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Summarize other answer and what I know all ways to do this using a one-liner      String testString    a b c d     1  Using Apache Commons  int apache   StringUtils countMatches testString        System out println  apache       apache     2  Using Spring Framework s  int spring   org springframework util StringUtils countOccurrencesOf testString        System out println  spring       spring     3  Using replace  int replace   testString length   - testString replace          length    System out println  replace       replace     4  Using replaceAll  case 1   int replaceAll   testString replaceAll             length    System out println  replaceAll       replaceAll     5  Using replaceAll  case 2   int replaceAllCase2   testString length   - testString replaceAll            length    System out println  replaceAll  second case        replaceAllCase2     6  Using split  int split   testString split       -1  length-1  System out println  split       split     7   Using Java8  case 1   long java8   testString chars   filter ch - gt  ch        count    System out println  java8       java8     8   Using Java8  case 2   may be better for unicode than case 1  long java8Case2   testString codePoints   filter ch - gt  ch        count    System out println  java8  second case        java8Case2     9    Using StringTokenizer  int stringTokenizer   new StringTokenizer      testString             countTokens  -1  System out println  stringTokenizer       stringTokenizer     From comment  Be carefull for the StringTokenizer  for a b c d it will work but for a   b c    d or    a b c d or a    b      c     d    or etc  it will not work  It just will count for   between characters just once  More info in github  Perfomance test  using JMH  mode   AverageTime  score 0 010 better then 0 351     Benchmark              Mode  Cnt  Score    Error  Units 1  countMatches        avgt    5  0 010     0 001  us op 2  countOccurrencesOf  avgt    5  0 010     0 001  us op 3  stringTokenizer     avgt    5  0 028     0 002  us op 4  java8 1             avgt    5  0 077     0 005  us op 5  java8 2             avgt    5  0 078     0 003  us op 6  split               avgt    5  0 137     0 009  us op 7  replaceAll 2        avgt    5  0 302     0 047  us op 8  replace             avgt    5  0 303     0 034  us op 9  replaceAll 1        avgt    5  0 351     0 045  us op

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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   gt 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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This is what I use to count the occurrences of a string  Hope someone finds it helpful      private long countOccurrences String occurrences  char findChar           return  occurrences chars   filter  x - gt                return x    findChar             count

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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

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I had an idea similar to Mladen  but the opposite     String s    a b c d   int charCount   s replaceAll             length    println charCount

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I had an idea similar to Mladen  but the opposite     String s    a b c d   int charCount   s replaceAll             length    println charCount

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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  gt   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

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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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Try this code   package com java test   import java util HashMap  import java util Map   public class TestCuntstring        public static void main String   args             String name    Bissssmmayaa           char   ar   new char name length             for  int i   0  i  lt  name length    i                  ar i    name charAt i                     Map lt Character  String gt  map new HashMap lt Character  String gt             for  int i   0  i  lt  ar length  i                  int count 0              for  int j   0  j  lt  ar length  j                      if ar i   ar j                        count                                                map put ar i   count   no of times                      System out println map

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String s    a b c d   long result   s chars   filter ch - gt  ch         count

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Using Java 8 and a HashMap without any library for counting all the different chars   private static void countChars String string        HashMap lt Integer  Integer gt  hm   new HashMap lt Integer  Integer gt         string chars   forEach letter - gt  hm put letter   hm containsKey letter    hm get letter    0    1        hm forEach  c  i  - gt  System out println   char c intValue            i

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Summarize other answer and what I know all ways to do this using a one-liner      String testString    a b c d     1  Using Apache Commons  int apache   StringUtils countMatches testString        System out println  apache       apache     2  Using Spring Framework s  int spring   org springframework util StringUtils countOccurrencesOf testString        System out println  spring       spring     3  Using replace  int replace   testString length   - testString replace          length    System out println  replace       replace     4  Using replaceAll  case 1   int replaceAll   testString replaceAll             length    System out println  replaceAll       replaceAll     5  Using replaceAll  case 2   int replaceAllCase2   testString length   - testString replaceAll            length    System out println  replaceAll  second case        replaceAllCase2     6  Using split  int split   testString split       -1  length-1  System out println  split       split     7   Using Java8  case 1   long java8   testString chars   filter ch - gt  ch        count    System out println  java8       java8     8   Using Java8  case 2   may be better for unicode than case 1  long java8Case2   testString codePoints   filter ch - gt  ch        count    System out println  java8  second case        java8Case2     9    Using StringTokenizer  int stringTokenizer   new StringTokenizer      testString             countTokens  -1  System out println  stringTokenizer       stringTokenizer     From comment  Be carefull for the StringTokenizer  for a b c d it will work but for a   b c    d or    a b c d or a    b      c     d    or etc  it will not work  It just will count for   between characters just once  More info in github  Perfomance test  using JMH  mode   AverageTime  score 0 010 better then 0 351     Benchmark              Mode  Cnt  Score    Error  Units 1  countMatches        avgt    5  0 010     0 001  us op 2  countOccurrencesOf  avgt    5  0 010     0 001  us op 3  stringTokenizer     avgt    5  0 028     0 002  us op 4  java8 1             avgt    5  0 077     0 005  us op 5  java8 2             avgt    5  0 078     0 003  us op 6  split               avgt    5  0 137     0 009  us op 7  replaceAll 2        avgt    5  0 302     0 047  us op 8  replace             avgt    5  0 303     0 034  us op 9  replaceAll 1        avgt    5  0 351     0 045  us op

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Complete sample   public class CharacterCounter      public static int countOccurrences String find  String string          int count   0      int indexOf   0       while  indexOf  gt  -1              indexOf   string indexOf find  indexOf   1         if  indexOf  gt  -1          count               return count          Call   int occurrences   CharacterCounter countOccurrences  l    Hello World     System out println occurrences      3

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

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