Compare Version Numbers

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:

1. Version strings are composed of numeric strings separated by dots . and this numeric strings may have leading zeroes.

2. Version strings do not start or end with dots, and they will not be two consecutive dots.

分析

.分开每个数字，一个个数字比较。记得拆开数字用Int(),可以去掉'01' = 》1

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('.')]
        for i in range(max(len(v1),len(v2))):
            n1 = v1[i] if i < len(v1) else 0
            n2 = v2[i] if i < len(v2) else 0
            if n1 > n2:
                return 1
            elif n1 < n2:
                return -1
        return 0