[LeetCode]Compare Version Numbers
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Compare two version numbers version1 and version1.
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
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
public class Solution { public int compareVersion(String version1, String version2) { String[] v1 = version1.split("\\."); String[] v2 = version2.split("\\."); for(String s:v1){ System.out.println(s); } for(int i=0;i<v1.length&&i<v2.length;i++){ int n1 = Integer.parseInt(v1[i]),n2 = Integer.parseInt(v2[i]); if(n1>n2){ return 1; }else if(n1<n2){ return -1; } } if(v1.length<v2.length){ for(int i=v1.length;i<v2.length;i++){ if(Integer.parseInt(v2[i])>0) return -1; } }else if(v1.length>v2.length){ for(int i=v2.length;i<v1.length;i++){ if(Integer.parseInt(v1[i])>0) return 1; } } return 0; }}
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