for (int i = 0; i < s.length; i++) { Pattern pt = Pattern.compile("\\b\\d+.\\d+\\b"); Matcher matcher = pt.matcher(s[i]); while (matcher.find()) { res[i] = matcher.group(); } }

This is how I understood: an array is started that checks every line of the s[i] array. Method (do I call it correctly?) Pattern searches in the i-th line. Pattern.compile is, roughly speaking, a stencil of what we are looking for in a string (here are double ). pt.matcher(s[i]) compares the string with the stencil (?), and while (matcher.find()) {res[i] = matcher.group();} - as long as there are matches (i.e., found so far, writing to the res[i] array). I didn’t understand at all how the stencil works: \\b\\d+.\\d+\\b

Explain what is not said, please.

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1 answer 1

Method (do I call it correctly?) Pattern

class Pattern

Pattern.compile is, roughly speaking, a stencil of what we are looking for in a string (here are doubles).

compile itself is just a method, but in the arguments it really has a regular expression (or a template, although it is more correct to call it a regular expression)

You can read about regular expressions here (or in any Google issue of the corresponding request): http://en.wikibooks.org/wiki/%D0%A0%D0B5% D0% B3% D1% 83% D0% BB% D1% 8F% D1% 80% D0% BD% D1% 8B% D0% B5_% D0% B2% D1% 8B% D1% 80% D0% B0% D0% B6% D0% B5% D0% BD% D0% B8% D1% 8F

  • 2
    Specifically, this regularity is defined as: \ b - boundary, the word boundary \ d is a digit (or range 0-9) + is a quantifier indicating that the symbol is repeated 1 or more times. - any character except newline. Here, obviously, I meant just a period, but forgot to put the preceding escape slash. That is, the regularity is deciphered like this: word boundary / one or several numbers / any character / one or several numbers / word boundary - etki
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