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

Solution 1 Make the two string the same length, if one is not long as the other, add 0;

4ms 11.92%

public class Solution {    public int compareVersion(String version1, String version2) {        String[] str1=version1.split("\\.");      String[] str2=version2.split("\\.");         int length = Math.max(str1.length, str2.length);    for(int i = 0; i < length; i++){        int num1 = i < str1.length ? Integer.parseInt(str1[i]) : 0;        int num2 = i < str2.length ? Integer.parseInt(str2[i]) : 0;                if(num1 > num2){            return 1;        }else if(num1  < num2){            return -1;        }    }    return 0;    }}

Solution 2 Easier to understand

3ms 42.85

public class Solution {    public int compareVersion(String version1, String version2) {      String[] verArr1 = version1.split("\\.");        String[] verArr2 = version2.split("\\.");        int index1 = 0;        int index2 = 0;        while (index1 < verArr1.length || index2 < verArr2.length) {            int val1 = 0;            int val2 = 0;            if (index1 < verArr1.length) {                val1 = Integer.parseInt(verArr1[index1++]);            }            if (index2 < verArr2.length) {                val2 = Integer.parseInt(verArr2[index2++]);            }            if (val1 < val2) {                return -1;            }            if (val1 > val2) {                return 1;            }        }        return 0;    }}