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1 题目
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.
2 思路
好吧，这题很蛋疼，各种情况。有1.0与1比较的，有1.2.3与1.2比较的，考虑到各种情况，很复杂。还发现java里面split函数输入是正则表达式，要用"."符号分割的话，得这样.split"\\."，才能分割。
网上查了查，有一个思路是用递归解决的，http://www.tuicool.com/articles/3QV7NvV。完美解决 比较 1.0与1的问题。
我感觉实际操作中，会规定死版本的格式吧。如1.2.3与1.0.0。另外，我看实际中的一个代码，只要发现两个版本的字符串不一样，就提醒用户更新版本－ －。
3 代码
public int compareVersion(String version1, String version2) { if(version1.equals(version2)) return 0; int fversion1 , fversion2;//最左边字符串代表的数字 String sversion1,sversion2;//剔除.号左边的数字剩下的字符串 if(version1.contains(".")){ int pos = version1.indexOf("."); fversion1 = Integer.valueOf(version1.substring(0,pos)); sversion1 = version1.substring(pos+1,version1.length()); }else { fversion1 = Integer.valueOf(version1); sversion1 = "0";//预防比较1.0与1这种情况 } if(version2.contains(".")){ int pos = version2.indexOf("."); fversion2 = Integer.valueOf(version2.substring(0,pos)); sversion2 = version2.substring(pos+1,version2.length()); }else { fversion2 = Integer.valueOf(version2); sversion2 = "0"; } if(fversion1 > fversion2) return 1; else if(fversion1 < fversion2) return -1; else return compareVersion(sversion1, sversion2); }
[leet code 165]Compare Version Numbers
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原文地址：http://www.cnblogs.com/lingtingvfengsheng/p/4314352.html
