【LeetCode】165. Compare Version Numbers 解题报告（Python）
标签（空格分隔）： LeetCode
作者： 负雪明烛
id： fuxuemingzhu
个人博客： http://fuxuemingzhu.me/
题目地址：https://leetcode.com/problems/compare-version-numbers/description/
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.
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
题目大意
比较两个版本号的大小。
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
日期
2018 年 6 月 26 日 ———— 早睡早起
