import re

def has_hyperlink(answer: str) -> bool:
    pattern = r'(https?:\/\/[^\s)]+)|(\[[^\]]+\]\((https?:\/\/[^\s)]+)\))'
    return bool(re.search(pattern, answer))

 可抓到：
	•	https://example.com
	•	http://example.com/test
	•	[Google](https://google.com)