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.
Example 1:
Input: version1 = "0.1", version2 = "1.1"
Output: -1
Example 2:
Input: version1 = "1.0.1", version2 = "1"
Output: 1
Example 3:
Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1

// correct 
This code assumes that next level is zero if no mo levels in shorter version number. And than compare levels.

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;
}





// wrong 

// compare every level , if there is no numver , defaukt as 0 
class Solution {
    public int compareVersion(String version1, String version2) {
        // differecent cases 
        // charArray 
        // if the current char is a number, compare the two chars 
        // if one number is biger than the other, return , if not
        // move on to the next level, if the there is no char left , by deault, we will think the string 
        // which still has char wins 
        
        int len = Math.min(version1.length(), version2.length());
        for(int i = 0; i < len; i++){
            if(!Character.isDigit(version1.charAt(i))) continue;
            if(version1.charAt(i) - '0' > version2.charAt(i) - '0'){
                return 1;
            }else if(version1.charAt(i) - '0' < version2.charAt(i) - '0'){
                return -1;
            }else{
                continue;
            }
        }
        return version1.length() > version2.length() ? 1 : -1;
        
    }
}



42 / 72 test cases passed.
Status: Wrong Answer
Submitted: 0 minutes ago
Input:
"01"
"1"
Output:
-1
Expected:
0