Compare version Numbershtml
Compare two version numbers version1 and version2.
If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.java
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.git
Here is an example of version numbers ordering:express
0.1 < 1.1 < 1.2 < 13.37
Idea:api
Just use the appropriate method: String#split()
.oracle
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("."))
.app
my Version:ide
public class Solution {this
public int compareVersion(String version1, String version2) {spa
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;
}
}
純屬模仿別人的代碼