leetcode 165. Compare Version Numbers

最新推荐文章于 2022-07-18 13:37:52 发布

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最新推荐文章于 2022-07-18 13:37:52 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`.`分割成序列，并对比每个序列中的数字来判断版本号的大小。文章提供了Python实现示例，并通过具体例子展示了如何使用该算法。

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Compare two version numbers version1 and version2.
If version1 > version2 return1; 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, 2.5is not “two and a half” or “half way to version three”, it is the fifth second-level revision of the second first-level revision.

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

版本号的比较，我们可以认为是浮点数的比较。split之后，用长的字符串减去短的字符串，忽略其中的0元素。

class Solution(object):
    def compareVersion(self, version1, version2):
        """
        :type version1: str
        :type version2: str
        :rtype: int
        """
        v1 = [int(x) for x in version1.split(".")]
        v2 = [int(x) for x in version2.split(".")]

        diff = []
        for i in range(min(len(v1), len(v2))):
            diff.append(v1[i] - v2[i])

        if len(v1) > len(v2):
            diff += [x for x in v1[len(v2):]]
        else:
            diff += [-x for x in v2[len(v1):]]
        if all(x == 0 for x in diff):
            return 0

        for i in range(len(diff)):
            if diff[i] == 0:
                continue

            if diff[i] > 0:
                return 1
            else:
                return -1`