165. Compare Version Numbers - Medium
Compare two version numbersversion1andversion2.
Ifversion1>version2
return1;
ifversion1<version2
return-1;
otherwise return0
.
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 be0
. For example, version number3.4
has a revision number of3
and4
for its first and second level revision number. Its third and fourth level revision number are both0
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 stringsmayhave leading zeroes. - Version strings do not start or end with dots, and they will not be two consecutive dots.
time = O(n), space = O(n)
class Solution { public int compareVersion(String version1, String version2) { String[] v1 = version1.split("\\."); String[] v2 = version2.split("\\."); int n = Math.max(v1.length, v2.length); for(int i = 0; i < n; i++) { int x = i < v1.length ? Integer.parseInt(v1[i]) : 0; int y = i < v2.length ? Integer.parseInt(v2[i]) : 0; if(x > y) { return 1; } else if(x < y) { return -1; } else { continue; } } return 0; } }