1+ /*
2+ Author: King, wangjingui@outlook.com
3+ Date: Dec 15, 2014
4+ Problem: Compare Version Numbers
5+ Difficulty: Easy
6+ Source: https://oj.leetcode.com/problems/compare-version-numbers/
7+ Notes:
8+ Compare two version numbers version1 and version1.
9+ If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0.
10+
11+ You may assume that the version strings are non-empty and contain only digits and the . character.
12+ The . character does not represent a decimal point and is used to separate number sequences.
13+ 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.
14+
15+ Here is an example of version numbers ordering:
16+
17+ 0.1 < 1.1 < 1.2 < 13.37
18+
19+ Solution: ...
20+ */
21+
22+ public class Solution {
23+ public int compareVersion (String version1 ,String version2 ) {
24+ long a =0 ,b =0 ;
25+ int v1len =version1 .length (),v2len =version2 .length ();
26+ int i =0 ,j =0 ;
27+ while (i <v1len ||j <v2len ) {
28+ a =0 ;b =0 ;
29+ while (i <v1len &&version1 .charAt (i ) !='.' ) {
30+ a =a *10 +version1 .charAt (i ) -'0' ;
31+ ++i ;
32+ }
33+ ++i ;
34+ while (j <v2len &&version2 .charAt (j ) !='.' ) {
35+ b =b *10 +version2 .charAt (j ) -'0' ;
36+ ++j ;
37+ }
38+ ++j ;
39+ if (a >b )return 1 ;
40+ if (a <b )return -1 ;
41+ }
42+ return 0 ;
43+ }
44+ }
