165. Compare Version Numbers Leetcode Python

最新推荐文章于 2021-03-03 11:17:39 发布

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本文介绍了一种用于比较两个版本号大小的算法实现。通过将版本号字符串按`.`分割，并逐段对比数字大小来判断版本先后。针对不同长度的版本号，通过补0的方式统一比较基准。

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
Credits:

Special thanks to @ts for adding this problem and creating all test cases.

这题主要要考虑不等长度的情况。

我的做法是当不等长的时候在前面加0

class Solution:
    # @param version1, a string
    # @param version2, a string
    # @return an integer
    def compareVersion(self, version1, version2):
        num1=version1.split('.')
        num2=version2.split('.')
        while len(num1) or len(num2):
            if len(num1)==0:
                num1=[0]
            elif len(num2)==0:
                num2=[0]
            else:
                i1=int(num1[0])
                i2=int(num2[0])
                if i1<i2:
                    return -1
                elif i1>i2:
                    return 1
                else:
                    num1=num1[1:]
                    num2=num2[1:]
        return 0