Leetcode - Compare Version Numbers

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Question

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

Java Code

public int compareVersion(String version1, String version2) {    String[] s1 = null;    String[] s2 = null;    //从输入字符串中解析提取到版本号中的各段数字,存储在数组中    s1 = version1.contains(".") ? version1.split("[.]") : new String[]{version1};    s2 = version2.contains(".") ? version2.split("[.]") : new String[]{version2};    int len1 = s1.length;    int len2 = s2.length;    int maxLen = len1 > len2 ? len1 : len2;    int num1;    int num2;    //逐段比较数字的大小    for(int i = 0; i < maxLen; ++i) {        num1 = (i < len1) ? Integer.parseInt(s1[i]) : 0;        num2 = (i < len2) ? Integer.parseInt(s2[i]) : 0;        if(num1 > num2)            return 1;        else if(num1 < num2)            return -1;    }    return 0;   }

说明

  • 这里给出代码不区分两个版本号之间的长短关系,实际上是将其中较短的版本号的后几个字段用0补全,使两者的字段一样多,方便进行比较,但是在每次比较之前需要判断数组指针是否越界,所以性能较低。当然也可以只比较两者共有的前若干个字段,然后再单独将较长版本号的后几个字段和0做比较,但是这样实现起来代码就比较冗长啰嗦了。
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