Compare two version strings
A deploy tool decides whether the version on the server is behind the one being released.
- Versions are dot-separated numbers: 1.2.10 has parts 1, 2 and 10.
- Compare part by part numerically, so 1.10 is above 1.2 — not below it, as a text sort would have it.
- A missing part counts as zero, so 1.2 and 1.2.0 are the same version.
- Leading zeros mean nothing: 1.01 is 1.1.
- Return -1 when the first is older, 1 when it is newer, 0 when they match.
compareVersions(first: string, second: string) → int
Java needs a compiler and Drill does not host one yet, so this page is the reference rather than an exercise: the problem, worked examples, and the solution in full. To type it out, the same problem runs in Python.
Where you start
int compareVersions(String first, String second) {
}Worked examples
| Call | Result |
|---|---|
compareVersions("1.2.10", "1.10.2") | -1 |
compareVersions("1.2", "1.2.0") | 0 |
compareVersions("2.0", "1.9.9") | 1 |
compareVersions("1.0.0", "1.0.0") | 0 |
Hint
Walk both to the length of the longer one, reading a missing part as zero.
Reference solution in Java
int compareVersions(String first, String second) {
String[] a = first.split("\\."), b = second.split("\\.");
int n = Math.max(a.length, b.length);
for (int i = 0; i < n; i++) {
int x = i < a.length ? Integer.parseInt(a[i]) : 0;
int y = i < b.length ? Integer.parseInt(b[i]) : 0;
if (x != y) return x < y ? -1 : 1;
}
return 0;
}