[165. Compare Version Numbers(https://leetcode.com/problems/compare-version-numbers/) 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.
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
解题方法 版本是用.进行分割的，那么我们也只能通过用.进行分割来判定版本号的大小。把版本号进行分割，需要找出一个较长的版本号的长度，把较短的版本的后面用0进行补齐。恩，然后进行比较。 class Solution(object): def compareVersion(self, version1, version2): """ :type version1: str :type version2: str :rtype: int """ v1_split = version1.split('.') v2_split = version2.split('.') v1_len, v2_len = len(v1_split), len(v2_split) maxLen = max(v1_len, v2_len) for i in range(maxLen): temp1, temp2 = 0, 0 if i < v1_len: temp1 = int(v1_split[i]) if i < v2_len: temp2 = int(v2_split[i]) if temp1 < temp2: return -1 elif temp1 > temp2: return 1 return 0