Leetcode-165. Compare Version Numbers

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前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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

这个题目有一个注意的就是split点的时候，必须两次转义Your runtime beats 28.88% of java submissions.

public class Solution {    public int compareVersion(String version1, String version2) {        int i = 0, j = 0;        String[] version1s = version1.split("\\.");        String[] version2s = version2.split("\\.");        while( i < version1s.length && j < version2s.length){            if(Integer.parseInt(version1s[i]) > Integer.parseInt(version2s[j])) return 1;            else if(Integer.parseInt(version1s[i]) < Integer.parseInt(version2s[j])) return -1;            i++;j++;        }        while(i < version1s.length) {if(Integer.parseInt(version1s[i]) != 0) return 1;i++;}        while(j < version2s.length) {if(Integer.parseInt(version2s[j]) != 0) return -1;j++;}        return 0;    }}

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