Java Regex否定提前错误匹配

Question

I'm looking for strings where the two first digits are present (in any order) in the digits that follow the space character.First I tried

(\d)(\d)\s\d*(\1|\2)\d*[\1\2&&[^\3]][\d]*

but it seems that I can't use brackets with backreferences.I tried using the lookahead feature instead with

(\d)(\d)\s\d*(\1|\2)\d*(?!\3(\1|\2))\d*

but I isn't right.The idea was "look for two digits, followed by a space, followed by zero or more digits, followed by either of the captured digits, followed by zero or more digits, followed by one of the captured digits which ISN'T the one I got before, followed by zero or more digits".21 20329 is a match.Why?How do I look for the strings I need?

Answer 1

This is simpler.

^(\d)(\d) (?=.*?\1)(?=.*?\2)\d+

See demo

1. The first lookahead ensures that the digit captured by Group 1 is present somewhere later in the string.
2. The second lookahead ensures that the digit captured by Group 2 is present somewhere later in the string.
3. If these conditions are met, the \\d+ eats up all the digits after the space.