leetcode Compare Version Numbers

Compare two version numbers version1 and version1.
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.

题目还是简单的,但是要考虑很多情况。

我写的代码边写边调试,写的比较烂。

 1 public class Solution {
 2     public int compareVersion(String version1, String version2) {
 3         String[] versionsOne=version1.split("\.");
 4         String[] versionsTwo=version2.split("\.");
 5         int oneLength=versionsOne.length;
 6         int twoLength=versionsTwo.length;
 7         int length=oneLength<twoLength?oneLength:twoLength;
 8         
 9         for (int i = 0; i < length; i++) {
10             while (versionsOne[i].startsWith("0")&&versionsOne[i].length()>1) {
11                 versionsOne[i]=versionsOne[i].substring(1);
12             }
13             while (versionsTwo[i].startsWith("0")&&versionsTwo[i].length()>1) {
14                 versionsTwo[i]=versionsTwo[i].substring(1);
15             }
16             if (versionsOne[i].length()>versionsTwo[i].length()) {
17                 return 1;
18             }else if (versionsOne[i].length()<versionsTwo[i].length()) {
19                 return -1;
20             }
21             int l = versionsOne[i].length()<versionsTwo[i].length()?versionsOne[i].length():versionsTwo[i].length();
22             for (int j = 0; j < l; j++) {
23                 int a=versionsOne[i].charAt(j);
24                 int b=versionsTwo[i].charAt(j);
25                 if (a>b) {
26                     return 1;
27                 }else if (a<b) {
28                     return -1;
29                 }
30             }
31             
32         }
33         if (oneLength>twoLength) {
34             for (int i = twoLength; i < oneLength; i++) {
35                 for (int j = 0; j < versionsOne[i].length(); j++) {
36                     if (versionsOne[i].charAt(j)>'0') {
37                         return 1;
38                     }
39                 }
40             }
41         }else if (oneLength<twoLength) {
42             for (int i = oneLength; i < twoLength; i++) {
43                 for (int j = 0; j < versionsTwo[i].length(); j++) {
44                     if (versionsTwo[i].charAt(j)>'0') {
45                         return -1;
46                     }
47                 }
48             }
49         }
50         return 0;
51     }
52 }
原文地址:https://www.cnblogs.com/birdhack/p/4168180.html