Compare Version Number

题目描述

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.

Here is an example of version numbers ordering:

0.1 < 1.1 < 1.2 < 13.37

解题方法

这一个题目并不难，但是需要考虑各种case, 所以在面试的时候要跟考官交流好可能的version number有哪些.

比如会出现的有

• 01
• 0.1.2
• 01.2.1

所以不一定只有一个小数点，在split之后，应该取长度更长的array来Loop，如果已经超出array长度就取0来比较

Solution

class Solution(object):
    def compareVersion(self, version1, version2):
        """
        :type version1: str
        :type version2: str
        :rtype: int
        """
        if not version1 or not version2:
            return 
        v1 = version1.split(".")
        v2 = version2.split(".")

        length = max(len(v1), len(v2)) # 取v1, v2长度更长的那个
        for i in range(length):
            # 如果没有了就用0来比较
            num1 = int(v1[i]) if i < len(v1) else 0
            num2 = int(v2[i]) if i < len(v2) else 0
            if num1 > num2:
                return 1
            elif num1 < num2:
                return -1
            else:
                continue

        return 0

Reference