165. Compare Version Numbers - Medium
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.
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; } }
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