其他分享
首页 > 其他分享> > leetcode 165. Compare Version Numbers

leetcode 165. Compare Version Numbers

作者:互联网

165. Compare Version Numbers
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.

比价两个版本号的大小

解题思路

  1. 将string以"."隔断拆分成list,
  2. 短的list的索引范围内逐个比较两个list中元素的大小
  3. 还未比出大小的情况下, 说明两个list在一定范围内都相等, 但是(1) 可能存在长list长出的部分全为0(说明两个list大小相等). (2) 可能存在若干个0但是末尾元素大于0(说明此时长list一定大于短list).
  4. 时间 空间复杂度均为O(n)

代码
Python

class Solution:
    def compareVersion(self, version1, version2):
        
        version1 = [int(_) for _ in version1.split(".")]
        version2 = [int(_) for _ in version2.split(".")]
        # print(version1, version2)
        len1 = len(version1)
        len2 = len(version2)
        
        for i in range(min(len1, len2)):
            if version1[i] > version2[i]:
                return 1
            elif version1[i] < version2[i]:
                return -1
        diff = len1 - len2
        if diff > 0 and version1[-1] != 0:
            return 1
        elif diff < 0 and version2[-1] != 0:
            return -1
        else:
            return 0

标签:Compare,return,version1,version2,list,len1,Version,diff,165
来源: https://blog.csdn.net/Bean771606540/article/details/100041831