Leetcode:compare Version Numbers

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, was not "both and 2.5 a half" or "half-to-version three", it is the fifth Second-level revision of the S Econd first-level revision.

Here are an example of version numbers ordering:

0.1 < 1.1 < 1.2 < 13.37

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

Approximate test instructions: Compare the version number, the two elements of the comparison are composed of a comma split string

intCompareversion (stringVersion1,stringVersion2) {Version1+='.'; Version2+='.'; intL1=version1.length (), l2=version2.length (), i,j,num1,num2; I=0; J=0;  while(I<l1 | | j<L2) {NUM1=0;  while(I<l1 && version1[i]!='.') {NUM1=num1*Ten+version1[i]-'0'; ++i; }        if(version1[i]=='.') ++i; Num2=0;  while(J<l2 && version2[j]!='.') {num2=num2*Ten+version2[j]-'0'; ++J; }        if(version2[j]=='.') ++J; if(num1!=num2)returnNum1>num2?1:-1; }    return 0; }

The interesting thing about this problem is the unified approach.

First in the string Wingarton, so that each time can be comma as a criterion

Second while (i<l1 | | j<L2) This sentence is also very interesting, I began to write the while (I<l1 && j<l2), In this case, each time the string length is different to determine the processing after the while

if return -1;     Else if return 1;

Because the front are all the same, not over the string affirmation version of the new, so glorious WA, there is data is 1 1.0,23333

So at the end of a string, whether or not the string is 0, but also to see, here just need to change the cycle conditions on the line

Leetcode:compare Version Numbers