The challenge
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.
You may assume the default revision number for each level of a version number to be 0
. For example, version number 3.4
has a revision number of 3
and 4
for its first and second level revision number. Its third and fourth level revision number are both 0
.
Example 1:
Input:version1
= "0.1",version2
= "1.1" Output: -1
Example 2:
Input:version1
= "1.0.1",version2
= "1" Output: 1
Example 3:
Input:version1
= "7.5.2.4",version2
= "7.5.3" Output: -1
Example 4:
Input:version1
= "1.01",version2
= "1.001" Output: 0 Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”
Example 5:
Input:version1
= "1.0",version2
= "1.0.0" Output: 0 Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"
Note:
- Version strings are composed of numeric strings separated by dots
.
and this numeric strings may have leading zeroes. - Version strings do not start or end with dots, and they will not be two consecutive dots.
The solution in Java
class Solution { public int compareVersion(String version1, String version2) { String[] arr1 = version1.split("\\."); String[] arr2 = version2.split("\\."); int i=0; while(i<arr1.length || i<arr2.length){ if(i<arr1.length && i<arr2.length){ if(Integer.parseInt(arr1[i]) < Integer.parseInt(arr2[i])) return -1; else if(Integer.parseInt(arr1[i]) > Integer.parseInt(arr2[i])) return 1; } else if(i<arr1.length){ if(Integer.parseInt(arr1[i]) != 0) return 1; } else if(i<arr2.length){ if(Integer.parseInt(arr2[i]) != 0) return -1; } i++; } return 0; } }