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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
该题的主要意图在于考察思考问题的全面性，该题体现了C/C++的性能优势，java代码贴出.
public class Solution { public int compareVersion(String version1, String version2) { String[] sv1 = version1.split("\\."); String[] sv2 = version2.split("\\."); int idx1 = 0; int idx2 = 0; while(idx1 < sv1.length || idx2 < sv2.length){ int v1 = 0; if(idx1 < sv1.length){ v1 = Integer.parseInt(sv1[idx1]); idx1++; } int v2 = 0; if(idx2 < sv2.length){ v2 = Integer.parseInt(sv2[idx2]); idx2++; } if(v1 < v2) return -1; if(v1 > v2) return 1; } return 0; } }
[LeetCode] Compare Version Numbers
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原文地址：http://www.cnblogs.com/HackingProgramer/p/4354757.html
