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Kotlin-based LeetCode algorithm problem solutions, regularly updated
165\. Compare Version Numbers
Medium
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**.