---
description: 'Author: @wingkwong | https://leetcode.com/problems/compare-version-numbers/'
---

# 0165 - Compare Version Numbers (Medium)

## Problem Link

https://leetcode.com/problems/compare-version-numbers/

## Problem Statement

Given two version numbers, `version1` and `version2`, compare them.

Version numbers consist of **one or more revisions** joined by a dot `'.'`. Each revision consists of **digits** and may contain leading **zeros**. Every revision contains **at least one character**. Revisions are **0-indexed from left to right**, with the leftmost revision being revision 0, the next revision being revision 1, and so on. For example `2.5.33` and `0.1` are valid version numbers.

To compare version numbers, compare their revisions in **left-to-right order**. Revisions are compared using their **integer value ignoring any leading zeros**. This means that revisions `1` and `001` are considered **equal**. If a version number does not specify a revision at an index, then **treat the revision as `0`**. For example, version `1.0` is less than version `1.1` because their revision 0s are the same, but their revision 1s are `0` and `1` respectively, and `0 < 1`.

_Return the following:_

* If `version1 < version2`, return `-1`.
* If `version1 > version2`, return `1`.
* Otherwise, return `0`.

**Example 1:**

```
Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both "01" and "001" represent the same integer "1".
```

**Example 2:**

```
Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: version1 does not specify revision 2, which means it is treated as "0".
```

**Example 3:**

```
Input: version1 = "0.1", version2 = "1.1"
Output: -1
Explanation: version1's revision 0 is "0", while version2's revision 0 is "1". 0 < 1, so version1 < version2.
```

**Constraints:**

* `1 <= version1.length, version2.length <= 500`
* `version1` and `version2` only contain digits and `'.'`.
* `version1` and `version2` **are valid version numbers**.
* All the given revisions in `version1` and `version2` can be stored in a **32-bit integer**.

## Approach 1: Split

We split the input strings by "." and store the digit into a list. Then we pad 0s to the shortest list to make the length for both list equal. Then we can compare both list. In Python3, $$cmp(a, b)$$ is gone so the equivalent for it is $$(a > b) - (a < b)$$.

<SolutionAuthor name="@wingkwong"/>

```python
class Solution:
    def compareVersion(self, version1: str, version2: str) -> int:
        v1 = [int(i) for i in version1.split(".")]
        v2 = [int(i) for i in version2.split(".")]
        l1, l2 = len(v1), len(v2)
        if l1 < l2: v1 += [0] * (l2 - l1)
        else: v2 += [0] * (l1 - l2)
        return (v1 > v2) - (v1 < v2)
```

