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

主要问题输入字符串有多种情况

  1. 不包含.的字符串

  2. 包含多个点的字符串

  3. 包含点的字符串后存在0的后缀,例如1.0.0与1的比较


对于字符串中.操作,在使用split切割字符串时要注意需要加“\\”

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;  


    }

}


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