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]; // 034556Note 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
