165 Compare Version Numbers

Approach 1: Three pass solution

直接想到的解, str.split('.') 然后比较每个数字, linear scan了三次

• version1 scan 1次 costs O(m)
• version2 scan 1次 costs O(n)
• compare 1次 costs O(max(m,n))

Note

\(O(max(n,m))\) in time, \(O(m+n)\) in space

class Solution:
    def compareVersion(self, version1: str, version2: str) -> int:
        one = version1.split('.')
        two = version2.split('.')


        for i in range(max(len(one),len(two))):
            left = 0 if i >= len(one) else int(one[i])
            right = 0 if i >= len(two) else int(two[i])

            if left > right:
                return 1

            if left < right:
                return -1

        return 0