^[1]*(0[1]*0[1]*)*$
which can be interpreted in the following way:
find any number of 1
followed by pairs of 00
with optional one or more 1
s in between.
More Related Contents:
- What do the characters ((?:(?:(?: mean in pattern-matching? [duplicate]
- The main thing is: how that binary for example 11110000 is divided into two parts such as s1=1111 and s2=0000. I have to do this in Java
- Program is saying all my valid inputs are invalid (REGEX code Issue maybe?) [closed]
- How to split a string if any chracter is found in java
- How to escape text for regular expression in Java
- Split string to equal length substrings in Java
- Why in Java 8 split sometimes removes empty strings at start of result array?
- Unicode equivalents for \w and \b in Java regular expressions?
- Difference between regex [A-z] and [a-zA-Z]
- Using Regular Expressions to Extract a Value in Java
- Java RegEx meta character (.) and ordinary dot?
- Getting the text that follows after the regex match
- How can I count the number of matches for a regex?
- What is the regex to extract all the emojis from a string?
- How to remove special characters from a string?
- Splitting string with pipe character (“|”) [duplicate]
- Replace the last part of a string
- Java equivalent to PHP’s preg_replace_callback
- java.lang.StackOverflowError while using a RegEx to Parse big strings
- Converting an int to a binary string representation in Java?
- Split regex to extract Strings of contiguous characters
- Java regex – overlapping matches
- Find everything between two XML tags with RegEx
- Parsing CSV input with a RegEx in java
- simple java regex throwing illegalstateexception [duplicate]
- Range of valid character for a base 64 encoding
- Regex: ?: notation (Question mark and colon notation) [duplicate]
- Java Pattern Matcher: create new or reset?
- How can I perform a partial match with java.util.regex.*?
- Java replace all square brackets in a string