165. Compare Version Numbers
Two Pointers, String ·Problem Statement
link: https://leetcode.com/problems/compare-version-numbers/ https://leetcode.cn/problems/compare-version-numbers/
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:
Input: ` version1 = “1.01”, version2 = “1.001”
**Output:** 0`
Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Input: version1 = "0.1", version2 = "1.1"
Output: -1
Solution Approach
The solution splits version strings into individual components, converts them to integers, and performs a straightforward comparison.
Algorithm
- Splitting: Break version1 and version2 on dots (‘.’).
- Comparison: Convert components to integers, using 0 for absent values, and compare.
- Result: Return -1, 1, or 0 based on the comparison outcome.
Implement
class Solution:
def compareVersion(self, version1: str, version2: str) -> int:
v1 = version1.split('.')
v2 = version2.split('.')
l1, l2 = len(v1), len(v2)
maxl = max(l1, l2)
for i in range(maxl):
p1 = int(v1[i]) if i < l1 else 0
p2 = int(v2[i]) if i < l2 else 0
if p1 < p2:
return -1
elif p1 > p2:
return 1
return 0