import re
octet = r'(?:25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)'
pattern=re.compile(r"\b(?!10\.|192\.168\.|172\.(?:1[6-9]|2[0-9]|3[01])\.){0}(?:\.{0}){{3}}\b".format(octet))
text = "10.11.0.0 and here are 192.168.0.0 and 192.168.0.2 145.12.24.45"
print(pattern.findall(text))
aW1wb3J0IHJlCm9jdGV0ID0gcicoPzoyNVswLTVdfDJbMC00XVswLTldfFswMV0/WzAtOV1bMC05XT8pJwpwYXR0ZXJuPXJlLmNvbXBpbGUociJcYig/ITEwXC58MTkyXC4xNjhcLnwxNzJcLig/OjFbNi05XXwyWzAtOV18M1swMV0pXC4pezB9KD86XC57MH0pe3szfX1cYiIuZm9ybWF0KG9jdGV0KSkKdGV4dCA9ICIxMC4xMS4wLjAgYW5kIGhlcmUgYXJlIDE5Mi4xNjguMC4wIGFuZCAxOTIuMTY4LjAuMiAxNDUuMTIuMjQuNDUiCnByaW50KHBhdHRlcm4uZmluZGFsbCh0ZXh0KSkK