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Counting the number of digits in an integer string in Java

I have a char array (say size 4). I am converting it to string using:

String str = String.valueOf(cf); // where cf is the char array of size 4.

This string can contain 01

either 11

or 001

or 011

etc. Now I need to calculate the number of digits in this line. But every time I calculate the number of digits (preferably in Java) the result is 4 (possibly because of the size of 4). How to calculate the value of no. digits according to the input line?

Example: if 001 is an input it should give o / p as 3, etc.

Here's the coding part:

static long solve(int k, long n)
    {

  //  System.out.println("Entered Solve function");
  char[] c = new char[4];

  long sum = 0;
  char[] cf = {};

  for(long i=2;i<=n;i++)
  {

      cf = fromDeci(c, k, i);
      String str = String.valueOf(cf);


       //System.out.println(snew);
       sum = sum + str.length() ;
    }

  return sum;
}
+3


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4 answers


You can use Stream API from java 8, solution in ONE-LINE:



String s = "101";
int nb = Math.toIntExact(s.chars()                     // convert to IntStream
                          .mapToObj(i -> (char)i)      // convert to char
                          .filter(ch -> isDigit(ch))   // keep the ones which are digits
                          .count());                   // count how any they are
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Very simple with regex:



String test = "a23sf1";
int count = 0;
Pattern pattern = Pattern.compile("[0-9]");
Matcher  matcher = pattern.matcher(test);
while (matcher.find()) {
    count++;
}
System.out.println(count);
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You can check if a character is a digit by simple comparison to digit characters.

private int countDigits(char[] cf) {
    int digitCt = 0; 
    for (char c : cf) {
        if ((c >= '0') && (c <= '9')) digitCt++;
    }
    return digitCt;
}
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Otherwise, you can use Apache Commons: (see StringUtils )

For example: StringUtils.getDigits("abc56ju20").size()

This gives you 4.

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