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_165_CompareVersionNumbers.java
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_165_CompareVersionNumbers.java
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package leetcode_1To300;
/**
* 本代码来自 Cspiration,由 @Cspiration 提供
* 题目来源:http://leetcode.com
* - Cspiration 致力于在 CS 领域内帮助中国人找到工作,让更多海外国人受益
* - 现有课程:Leetcode Java 版本视频讲解(1-900题)(上)(中)(下)三部
* - 算法基础知识(上)(下)两部;题型技巧讲解(上)(下)两部
* - 节省刷题时间,效率提高2-3倍,初学者轻松一天10题,入门者轻松一天20题
* - 讲师:Edward Shi
* - 官方网站:https://cspiration.com
* - 版权所有,转发请注明出处
*/
public class _165_CompareVersionNumbers {
/**
* 165. Compare Version Numbers
* 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
time : O(max(m,n))
space : O(n)
* @param version1
* @param version2
* @return
*/
public int compareVersion(String version1, String version2) {
String[] v1 = version1.split("\\.");
String[] v2 = version2.split("\\.");
for (int i = 0; i < Math.max(v1.length, v2.length); i++) {
int num1 = i < v1.length ? Integer.parseInt(v1[i]) : 0;
int num2 = i < v2.length ? Integer.parseInt(v2[i]) : 0;
if (num1 < num2) {
return -1;
} else if (num1 > num2) {
return 1;
}
}
return 0;
}
}