Leetcode#165Compare Version Numbers

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

You may assume that the version strings are non-empty and contain only digits and the . character.
The . character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.

Here is 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.

165

主要问题

1. 字符串中是否包含.字符

2. 字符串中包含多个.字符

3. 包含.字符的字符串后缀都是0的情况，例如1.0.0与1的比较

java字符串在处理.字符时，一定要注意添加"\\"->"\\."

public class Solution {

    public int compareVersion(String version1, String version2) {

     String w1[]=version1.split("\\.");

        String w2[]=version2.split("\\.");

        int l1=w1.length;

        int l2=w2.length;

        int l=l1;

        if(l2<l1)

            l=l2;

        int i=0;

        for(i=0;i<l;i++)

        {

            int a=Integer.parseInt(w1[i]);

            int b=Integer.parseInt(w2[i]);

            if(a<b)

                return -1;

            if(a>b)

                return 1;

        }

        if(l1<l2)

        {

         for(i=l;i<l2;i++)

         if(Integer.parseInt(w2[i])>0)

         return -1;

         return 0;

        }

        else if(l1>l2)

        {

         for(i=l;i<l1;i++)

         if(Integer.parseInt(w1[i])>0)

         return 1;

         return 0;

        }

        else

            return 0;  

    }

}