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
C++ 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(std::string first, std::string second) {
}Worked examples
| Call | Result |
|---|---|
compareVersions(std::string("1.2.10"), std::string("1.10.2")) | -1 |
compareVersions(std::string("1.2"), std::string("1.2.0")) | 0 |
compareVersions(std::string("2.0"), std::string("1.9.9")) | 1 |
compareVersions(std::string("1.0.0"), std::string("1.0.0")) | 0 |
Hint
Walk both to the length of the longer one, reading a missing part as zero.
Reference solution in C++
int compareVersions(std::string first, std::string second) {
auto parts = [](const string& s) {
std::vector<int> out;
string cur;
for (char c : s) {
if (c == '.') { out.push_back(cur.empty() ? 0 : std::stoi(cur)); cur.clear(); }
else cur += c;
}
out.push_back(cur.empty() ? 0 : std::stoi(cur));
return out;
};
auto a = parts(first), b = parts(second);
size_t n = std::max(a.size(), b.size());
for (size_t i = 0; i < n; i++) {
int x = i < a.size() ? a[i] : 0;
int y = i < b.size() ? b[i] : 0;
if (x != y) return x < y ? -1 : 1;
}
return 0;
}