难度: Medium
原题连接
内容描述
Compare two version numbers version1 and version2.
If version1 > version2 return 1; if version1 < version2 return -1;otherwise return 0.
You may assume that the version strings are non-empty and contain only digits and the . character.
The . character does not represent a decimal point and is used to separate number sequences.
For instance, 2.5 is not "two and a half" or "half way to version three", it is the fifth second-level revision of the second first-level revision.
You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.
Example 1:
Input: version1 = "0.1", version2 = "1.1"
Output: -1
Example 2:
Input: version1 = "1.0.1", version2 = "1"
Output: 1
Example 3:
Input: version1 = "7.5.2.4", version2 = "7.5.3"
Output: -1
Example 4:
Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”
Example 5:
Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"
Note:
Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.
Version strings do not start or end with dots, and they will not be two consecutive dots.
思路 1 - 时间复杂度: O(N)- 空间复杂度: O(N)******
class Solution:
def compareVersion(self, version1, version2):
"""
:type version1: str
:type version2: str
:rtype: int
"""
lst1 = [int(v) for v in version1.split('.')]
lst2 = [int(v) for v in version2.split('.')]
for i in range(max(len(lst1), len(lst2))):
num1 = lst1[i] if i < len(lst1) else 0
num2 = lst2[i] if i < len(lst2) else 0
if num1 > num2:
return 1
elif num1 < num2:
return -1
return 0