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Leetcode 165. Compare Version Numbers

2024-11-03 20:35:39   208人阅读

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

 

Next challenges: Text Justification Edit Distance Longest Uncommon Subsequence I

 

思路：按照"."先将version1和version2分别分割为字符串数组，然后一段一段比较，注意对于当前的一段可能会有前缀0的出现，比如"001"和"1"。另外将每一段转换为整数来处理是不可行的，比如"2.10000000000000000000000000000"和"2.1"。

 

代码：

 1 public int compareVersion(String version1, String version2) {
 2         String[] array1 = version1.split("\\.");
 3         String[] array2 = version2.split("\\.");
 4         int time = Math.max(array1.length, array2.length);
 5         for(int i = 0; i < time; i++) {
 6             String s1 = i < array1.length ? array1[i] : "0";
 7             String s2 = i < array2.length ? array2[i] : "0";
 8             //去掉前缀0
 9             int j = 0;
10             while(j < s1.length() - 1 && s1.charAt(j) == ‘0‘) j++;
11             if(j > 0) s1 = s1.substring(j);
12             j = 0;
13             while(j < s2.length() - 1 && s2.charAt(j) == ‘0‘) j++;
14             if(j > 0) s2 = s2.substring(j);
15             
16             if(s1.length() == s2.length()) {
17                 for(int k = 0; k < s1.length() ; k++) {
18                     if(s1.charAt(k) == s2.charAt(k)) continue;
19                     return s1.charAt(k) > s2.charAt(k) ? 1 : -1;                    
20                 }
21             }else{
22                 return s1.length() > s2.length() ? 1 : -1;
23             }
24         }
25         return 0;
26     }

 

Leetcode 165. Compare Version Numbers


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