[Java]Leetcode165 Compare Version Numbers

標籤：leetcode   string   

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

題意是：比較版本號碼的大小，假設版本號碼只含有數字和小數點。這一題首先可能想到的是字元轉換為整型比較大小，但是版本號碼中可以有多個小數點，這一點就不符合了。於是想到用split()方法，以小數點為分界符，分成一個字元數組，再一一比較。

代碼如下：

 public int compareVersion(String version1, String version2) {        String[] ver1=version1.split("\\.");        String[] ver2=version2.split("\\.");        int len1=ver1.length;        int len2=ver2.length;        int len=0;        if(len2<=len1)len=len2;        else len=len1;        int num1=0;        int num2=0;        int i=0;        while(i<len)        {           num1=Integer.valueOf(ver1[i]);           num2=Integer.valueOf(ver2[i]);           if(num1==num2)i++;           else if(num1<num2)return -1;           else return 1;        }        while(len1>i)//有小數點，但是小數點後的值為0的情況，不為0返回1，都為0就return 0.        {          if(Integer.valueOf(ver1[i++])!=0)return 1;        }        i=len;        while(len2>i)        {           if(Integer.valueOf(ver2[i++])!=0)return -1;          }        return 0;    }

[Java]Leetcode165 Compare Version Numbers