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 . characte…

Compare two version numbers version1 and version1.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 . characte…

原题链接在这里:https://leetcode.com/problems/compare-version-numbers/ 用string.split()方法把原有string 从小数点拆成 string 数组,但这里要注意 . 和 * 是不能直接用split(".") 或者split("*")拆开的,因为 . 可以代表任意char, * 可以代表任意字符串.所以要加 \\. 来避免individual special character. 拆开后用Interge…

[leetcode 字符串处理]Compare Version Numbers @author:wepon @blog:http://blog.csdn.net/u012162613 1.题目 Compare two version numbers version1 and version1. If version1 > version2 return 1, if version1 < version2 return -1, otherwise return 0. You may assume…

[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 vers…

Question 165. Compare Version Numbers Solution 题目大意: 比较版本号大小 思路: 根据逗号将版本号字符串转成数组,再比较每个数的大小 Java实现: public int compareVersion(String version1, String version2) { String[] v1Arr = version1.split("\\."); String[] v2Arr = version2.split("\\.&qu…

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 an…

Compare Version Numbers 本题收获: 1.字符串型数字转化为整型数字的方法:s[i] - '0',( 将字母转化为数字是[i]-'A'   ) 2.srt.at(),substr(),atoi() 3.字符串型数组的定义:string str[3]={“aaaa”,"bbbb","cccc"};差点跟python混淆了.C++中双引号和单引号的区别. 4.返回值类型要对应 return "0", 5.return 0:的位置…

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 . characte…

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 . characte…