There are many nice solution in iterative way, but not a recursive one, so here it is if you are interested.

It is longer than those nice iterative solutions, but `can be faster`

. Since those nice solutions split the two arrays, which take O(m+n) time for sure. Here, we are only looking at the leftmost part and stop whenever the comparison result is either -1 or 1.

```
public class Solution {
public int compareVersion(String version1, String version2) {
if(version1.euqals("") && version2.equals("")) return 0; // both are "", return 0
String sub1, sub2;
int v1s, v2s;
if(version1.equals("")) {
v1s = 0;
sub1 = "";
}
else {
int id1 = version1.indexOf(".");
if(id1==-1) {
v1s = Integer.parseInt(version1);
sub1 = "";
}
else {
v1s = Integer.parseInt(version1.substring(0, id1));
sub1 = version1.substring(id1+1);
}
}
if(version2.equals("")) {
v2s = 0;
sub2 = "";
}
else {
int id2 = version2.indexOf(".");
if(id2==-1) {
v2s = Integer.parseInt(version2);
sub2 = "";
}
else {
v2s = Integer.parseInt(version2.substring(0, id2));
sub2 = version2.substring(id2+1);
}
}
if(v1s>v2s) return 1;
if(v1s<v2s) return -1;
return compareVersion(sub1, sub2);
}
}
```

BTW, is there any way to simplify it?