leetcode Compare Version Numbers

Compare two version numbers version1 and version1.
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


解法：

2，采用“.”分离时，要加上“\.”
3，数组没有赋值时，默认为0

package Leetcode;

public class CompareVersionNumbers {
    public int compareVersion(String version1, String version2) {
        String[] ver1=version1.split("\.");
        String[] ver2=version2.split("\.");
        int maxlength=Math.max(ver1.length, ver2.length);
        
        int[] ints1 = new int[maxlength];
        
        for (int i=0; i < maxlength; i++) {
            ints1[i] = Integer.parseInt(ver1[i]);
        }
        int[] ints2 = new int[maxlength];
        for (int i=0; i < maxlength; i++) {
            ints2[i] = Integer.parseInt(ver2[i]);
        }
        for(int i=0;i<maxlength;i++){
            if(ints1[i]>ints2[i]){
                return 1;
            }
            else if(ints1[i]<ints2[i]){
                return -1;
            }
        }
        return 0;
    }
}

 2017/8/19 更新， clean code

class Solution {
    public int compareVersion(String version1, String version2) {
        String[] ints1 = version1.split("\.");
    String[] ints2 = version2.split("\.");
    
    int length = Math.max(ints1.length, ints2.length);
    for (int i=0; i<length; i++) {
        Integer v1 = i < ints1.length ? Integer.parseInt(ints1[i]) : 0;
        Integer v2 = i < ints2.length ? Integer.parseInt(ints2[i]) : 0;
        int compare = v1.compareTo(v2);
        if (compare != 0) {
            return compare;
        }
    }
    
    return 0;
    }
}