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Java [Leetcode 165]Compare Version Numbers

2016-02-22 14:54 627 查看
题目描述:

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

解题思路:

以小数点分开,逐个比较。

代码如下:

public class Solution {
public int compareVersion(String version1, String version2) {
String[] levels1 = version1.split("\\.");
String[] levels2 = version2.split("\\.");
int length = Math.max(levels1.length, levels2.length);
for(int i = 0; i < length; i++){
Integer v1 = i < levels1.length ? Integer.parseInt(levels1[i]) : 0;
Integer v2 = i < levels2.length ? Integer.parseInt(levels2[i]) : 0;
int compare = v1.compareTo(v2);
if(compare != 0)
return compare;
}
return 0;
}
}


  
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