Java [Leetcode 165]Compare Version Numbers

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;
    }
}

Java [Leetcode 165]Compare Version Numbers

原文：http://www.cnblogs.com/zihaowang/p/5207128.html