Huawei OJ Test question--determine if the input string is a valid IP address (with the most complete test data in the world)

Topic title:

• Determines whether the input string is a valid IP address

Detailed Description:

Please implement the following interface

< Span style= "margin:0px; padding:0px; Font-family: new song body; Color:blue; font-size:12px ">booli sipaddressvalid (const char * pszipaddr "

Input: pszipaddr string

Output: True valid IP address, false, invalid IP address

Constraints

2. Input IP is XXX.XXX.XXX.XXX format

4. The string ends with a space that is considered a legitimate IP

6. The middle of the string contains spaces that are considered illegal IPs

8. Similar to 01.1.1.1, the 1.02.3.4 IP sub-segment starts with 0 as an illegal IP

10. A single 0 is considered legitimate ip,0.0.0.0 is also counted as a legitimate IP

This topic is very simple, but our code can not be tested, for what, because this format requires a lot of default, we have to do careful people!

Give multiple sets of test data:

/**********************************

Null

0.0.0.0

01.2.0.6

255.2.256.0

256.22.234

34.45.345.6

2.3.3.3

0.1.0.0

00.00.32.67

**********************************/

Here is the code, Welcome to add!

#include "IPAddressValid.h" #include <iostream>using namespace std; #include "IPAddressValid.h"  #include <cstring>  #include <iostream>  #include <cstdio>  using namespace std;    BOOL Isipaddressvalid (const char* pszipaddr) {if (Pszipaddr==null | | pszipaddr== ") return false;    Please implement int Slen=strlen (PSZIPADDR) here;//cout<<slen<<endl;int I,j,k,m,n,sum;int Sublen[10];const Char *ss= Pszipaddr;<span style= "White-space:pre" ></span>

<span style= "White-space:pre" ></span>//remove the Kinsoku space for (I=0;i<slen && ss[i]== "; i++); if (I==slen) Return false;for (j=slen-1;j>=0 && ss[j]== "; j--); if (j==0) return false;if (I>=J) return false;

<span style= "White-space:pre" ></span>//determine if there are spaces in the middle of the string, and if so FalseFor (k=i;k<j;k++) if (ss[k]== ") return false;//cout<<i<< "" <<j<<endl;

<span style= "White-space:pre" ></span>//record string appears '. ' If the point appears in the first character or adjacent element is also point falsem=0;for (k=i;k<=j;k + +) {if (ss[k]== '. ') {sublen[m++]=k;if (k==i | | k==j) return false;if (ss[k+1]== '. ') return false;}}

<span style= "White-space:pre" ></span>//if the number of points is not 3 then Falseif (m!=3) return false;sublen[m++]=j+1;//for (k=1; k<m;k++)//if ((sublen[k]-sublen[k-1]) >4) return false;//for (k=0;k<m;k++)//cout<<sublen[k]<< " ";//cout<<endl;

<span style= "White-space:pre" ></span>//determine whether each sub-segment is 0, if it is the only 0 if the book is not, if not falsen=0;k=i;while (n<m) { For (K=i;k<sublen[n] && ss[k]== ' 0 '; k++)//if (ss[k]!= ' 0 ' && k<sublen[n] && k>i) Return false;if (ss[k]== ' 0 ') {if (K+1<slen && ss[k+1]!= '. ') return false;} k=sublen[n]+1;n++;}

<span style= "White-space:pre" ></span>//determine whether the integer conversion of each sub-paragraph is between 0~255, if not Falsen=0;while (n<m) {sum=0;for (k =i;k<sublen[n];k++) {if (ss[k]< ' 0 ' | | ss[k]> ' 9 ') return false;sum=sum*10+ss[k]-' 0 ';} if (sum>255) return false;i=sublen[n]+1;n++;}    return true;}

Huawei OJ Test question--determine if the input string is a valid IP address (with the most complete test data in the world)