문제
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.
version1, version2가 주어졌을 때 어떤게 최신 버전인지 찾는 문제이다.
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.
풀이
아이디어
- 버전은 앞자리가 높을수록 최신이다.
- 각 버전을 " . "을 기준으로 나누고 앞에서부터 비교한다.
코드
class Solution:
def compareVersion(self, version1: str, version2: str) -> int:
# '.' 을 기준으로 나눔
version1 = version1.split('.')
version2 = version2.split('.')
while version1 and version2:
v1 = version1.pop(0)
v2 = version2.pop(0)
if int(v1) > int(v2): return 1
elif int(v1) < int(v2): return -1
# .의 개수가 다른 경우 처리
if sum(map(int, version1)): return 1
if sum(map(int, version2)): return -1
return 0
후기
왜 Medium 난이도의 문제인지 잘 모르겠는 문제였다.
.split()의 사용법만 알고있으면 쉽고 편하게 풀 수 있을 것이다.