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.
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.
分析:1.注意 “.” 要转译“\.”
2. 主要consider many case 最简方法就是,没有就补0
public class Solution { public int compareVersion(String version1, String version2) { if(version1.equals(version2)) return 0; String[] v1 = version1.split("\."); String[] v2 = version2.split("\."); for(int i = 0 ; i < v1.length || i < v2.length ; i++){ String index1 = (i >= v1.length ? "0" : v1[i]); String index2 = (i >= v2.length ? "0" : v2[i]); if(convertStrtoInt(index1) > convertStrtoInt(index2)) return 1; else if (convertStrtoInt(index1) < convertStrtoInt(index2)) return -1; } return 0; } public int convertStrtoInt(String s){ int num = 0; for(int i = 0 ; i < s.length() ; i++){ num = num * 10 + s.charAt(i) - '0'; } return num; } }