Compare Version Number

Compare 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

Idea:

Just use the appropriate method: String#split().

String string = "004-034556";
String[] parts = string.split("-");
String part1 = parts[0]; // 004
String part2 = parts[1]; // 034556

Note that this takes a regular expression, so remember to escape special characters if necessary, e.g. if you want to split on period . which means "any character" in regex, use either split("\\.") or split(Pattern.quote(".")).

my Version:

public class Solution {

    public int compareVersion(String version1, String version2) {

        String[] v1 = version1.split("\\.");

        String[] v2 = version2.split("\\.");

        

        int length = Math.max(v1.length,v2.length);

        for(int i=0;i<length;i++){

            Integer s1 = i<v1.length?Integer.parseInt(v1[i]):0; //notice here must be a Integer Object

            Integer s2 = i<v2.length?Integer.parseInt(v2[i]):0; //otherwise cannot use the compareTo() 

            int result = s1.compareTo(s2);

            if(result != 0){

                return result;

            }

        }

        return 0;

    }

}

纯属模仿别人的代码