话那问题就简单了好多,这个实现效率比较低下.
boolean ipValid(String s){
String regex0="(2[0-4]\\d)" + "|(25[0-5])";
String regex1="1\\d{2}";
String regex2="[1-9]\\d";
String regex3="\\d";
String regex="("+regex0+")|("+regex1+")|("+regex2+")|("+regex3+")";
regex="("+regex+").("+regex+").("+regex+").("+regex+")";
Pattern p=Pattern.compile(regex);
Matcher m=p.matcher(s);
return m.matches();
}
延伸阅读
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