"Leetcode string Processing" Compare Version Numbers

"Leetcode string processing" Compare Version Numbers

@author: Wepon@blog: http://blog.csdn.net/u012162613
1. Topics

Compare numbers version1 and version1.
If version1 > version2 return 1, if version1 < version2 return-1, otherwise ret Urn 0.

Assume that the version strings is non-empty and contain only digits and the.Character.
The.Character does not represent a decimal point and was used to separate number sequences.
For instance,2.5is not "both and a half" or "half-to-version three", it is the fifth Second-level revision of the second first-level re Vision.

Here are an example of version numbers ordering:

0.1 < 1.1 < 1.2 < 13.37

2. Analysis Test Instructions: test instructions is very clear, is to compare the "version number" size, given the version number Version1 and Version2 is a string type, when Version1>version2, return 1, reverse 1.
This problem belongs to the details of the problem, in addition to string processing is not much more cumbersome.
problem-solving ideas: first, respectively, Version1, version2 string Press '. ' Split into multiple substrings, and each substring is converted into an integral type into a container. Finally, the size of the number of the corresponding position in the two containers is compared. Of course, they need to consider the different lengths of their cases.
Note the point:(1) Form of version: 10.23.1, 01.23, 1.2.3.4 ... and so forth .(2) string conversion to Integer, my program directly with the library function stoi ()
3. Code

Class Solution {Public:int compareversion (string version1, String version2) {vector<int> Result1=getint (        Version1);        Vector<int> Result2=getint (Version2);        int len1=result1.size ();        int len2=result2.size ();        if (LEN2&LT;LEN1) return-1*compareversion (Version2, Version1);        int i=0;                while (I<len1 && result1[i]==result2[i]) i++;            The IF (I==LEN1) {//STR1 and str2 len1 bits are equal, then look at whether the str2 bits behind Len2-len1 are all 0 to determine their size int j=len2-1;            while (J >= len1) {if (result2[j--]!=0) return-1;        } return 0;            }else{//str1 and str2 before the len1 position are not equal, directly determine the first I-bit if (result1[i]<result2[i]) return-1;        else return 1; }}private://the version string by '. '        Split into multiple, converted to integral type into the container vector<int> getInt (string version) {vector<int> result;        int len=version.size ();        int pre=0; for (int i=0;i<len;i++) {if (version[i]== '. ')      {          String str (Version.begin () +pre,version.begin () +i);                Note that this type of initialization, left closed right, that STR does not include Version[version.begin () +i] Result.push_back (Stoi (str));            pre=i+1;        }}string Str (Version.begin () +pre,version.end ());        Result.push_back (Stoi (str));    return result; }};

"Leetcode string Processing" Compare Version Numbers