165. compare version numbers

165. Compare version number

Original title link ：https://leetcode.cn/problems/compare-version-numbers/

Here are two version numbers version1 and version2 , Please compare them .

The version number consists of one or more revision numbers , Each revision number consists of a ‘.’ Connect . Each revision number consists of Multiple digits form , May contain Leading zero . Each version number contains at least one character . The revision number is numbered from left to right , Subscript from 0 Start , The leftmost revision number is subscript 0 , The next revision number is subscript 1 , And so on . for example ,2.5.33 and 0.1 All valid version numbers .

When comparing version numbers , Please compare their revision numbers from left to right . When comparing revision numbers , Just compare Ignore any integer values after leading zeros . in other words , Revision number 1 And revision number 001 equal . If the version number does not specify a revision number at a subscript , Then the amendment number shall be deemed to be 0 . for example , edition 1.0 Less than version 1.1 , Because they are subscript 0 The revision number of is the same , And the subscript is 1 The revision numbers of are respectively 0 and 1 ,0 < 1 .

The return rule is as follows ：

If version1 > version2 return 1,
If version1 < version2 return -1,
In addition, return to 0.

Example 1：

Input ：version1 = “1.01”, version2 = “1.001”
Output ：0
explain ： Ignore leading zeros ,“01” and “001” All represent the same integer “1”
Example 2：

Input ：version1 = “1.0”, version2 = “1.0.0”
Output ：0
explain ：version1 The subscript is not specified as 2 Revision number of , That is to say “0”
Example 3：

Input ：version1 = “0.1”, version2 = “1.1”
Output ：-1
explain ：version1 Subscript is 0 The revision number of is “0”,version2 Subscript is 0 The revision number of is “1” .0 < 1, therefore version1 < version2

Tips ：

1 <= version1.length, version2.length <= 500
version1 and version2 Contains only numbers and ‘.’
version1 and version2 All are Valid version number
version1 and version2 All revision numbers of can be stored in 32 An integer in

Their thinking ：

Compare each position of the string

Code implementation ：

class Solution:
    def compareVersion(self, version1: str, version2: str) -> int:
        from itertools import zip_longest
        for z1, z2 in zip_longest(version1.split('.'), version2.split('.'), fillvalue=0):
            #  After parsing, compare the values of the corresponding positions of the two version numbers 
            x, y = int(z1), int(z2)
            if x != y:
                return 1 if x > y else -1
        return 0

reference ：
https://leetcode.cn/problems/compare-version-numbers/solution/bi-jiao-ban-ben-hao-by-leetcode-solution-k6wi/