HW-IP legality _java

Describe	Now IPV4 is represented by a 32-bit unsigned integer, which is typically displayed in a point-and-click manner, dividing the IP address into 4 parts, each of which is 8 bits, which is represented as an unsigned integer (so there is no need to use a plus sign). As 10.137.17.1, we are very familiar with the IP address, an IP address string no space appears (because to be represented as a 32 number). Now you need to use the program to determine whether IP is legitimate.
Knowledge points	String, loop, list, queue, stack, find, search, sort, tree, graph, array, function, pointer, enumeration, bitwise operation, struct, union, file operation, recursive
Run time limit	10M
Memory limit	128
Input	Enter an IP address
Output	Returns the result of the decision Yes or NO
Sample input	10.138.15.1
Sample output	YES

Personal Summary:

(1)the use of regular expressions in Java syntax, for ', ' to use ' \ \ '

(2) Note that the empty string "" and the difference between NULL, determine whether a string is an empty string with. Equals ("")

(3) Judgment in the title itself

1. Note that the string array must be four in length, such as 1.1.1 is not legal
2. And you can't have an empty string, that is. Not adjacent, such as 1.1. 1 's not legal.
3. For each sub-string to determine whether the number is greater than or equal to 0, less than or equal to 255 of the numbers; I am using a single add, in fact, you can use Java Integer.parseint (string2);

(1, judging if there are three '. ') ; 2, determine whether the three points are not adjacent, 3, determine whether each part is a number, 4, determine whether each number is between 0 to 255. ）

Programme one:

ImportJava.util.Scanner; Public classMain { Public Static voidMain (String[]args) {Scanner Scanner=NewScanner (system.in); String String=Scanner.nextline (); string[] Num=string.split ("\ \")); Booleanresult =true; if(num.length!=4) result =false;  for(String string2:num) {intNo =Integer.parseint (string2); if(No>=0 && no<=255) {}Else{result=false;  Break; }}if(Result) {System.out.println ("YES"); }Else{System.out.println ("NO"); }　}
}

Scenario Two:

ImportJava.util.Scanner; Public classip_useful { Public Static voidMain (string[] args) {System.out.println ("Please enter the IP address according to the specification"); Scanner Scstr=NewScanner (system.in); String sqy=Scstr.nextline ();  System.out.println (judge (sqy)); //Print    }     Public Staticstring judge (String mxf) {string str[]=mxf.split ("\ \");//to separate a number of strings into dots        if(str.length!=4){            return"NO";//If there is not exactly three '. ' will return        }                intNum=0; intTmp=0; intFlag[] =New int[Str.length]; intFlag1=1;  for(inti=0;i<str.length;i++){            if(Str[i].equals ("")){                return"NO";//if an empty string is returned            }            Char[] Ch=str[i].tochararray ();//convert a single string to a character array            if(ch.length<=3) {//less than four digits to be effective                 for(intj=0;j<ch.length;j++){                    if(ch[j]>= ' 0 ' &&ch[j]<= ' 9 ') {//Digital Judgmenttmp=ch[j]-' 0 '; Num=num+tmp*10^ (ch.length-j-1); }                }                if(num>=0&&num<256) {Flag[i]=1; }            }        }         for(intk=0;k<str.length;k++) {Flag1=FLAG1*FLAG[K];//as long as each one of the mark bit is 1;        }                    if(flag1==1)           return"YES"; Else            return"NO"; }}

HW-IP legality _java