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.

You may assume the default revision number for each level of a version number to be 0. For example, version number 3.4 has a revision number of 3 and 4 for its first and second level revision number. Its third and fourth level revision number are both 0.

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

Example 4:

Input: version1 = "1.01", version2 = "1.001"
Output: 0
Explanation: Ignoring leading zeroes, both “01” and “001" represent the same number “1”

Example 5:

Input: version1 = "1.0", version2 = "1.0.0"
Output: 0
Explanation: The first version number does not have a third level revision number, which means its third level revision number is default to "0"

Solution 1:

class Solution {
public int compareVersion(String version1, String version2) {
if (version1 == null || version2 == null) {
return 0;
}
String[] strArr1 = version1.split("\\.");
String[] strArr2 = version2.split("\\.");
int index = 0;
while (index < strArr1.length && index < strArr2.length) {
int cur_str1 = Integer.parseInt(strArr1[index]);
int cur_str2 = Integer.parseInt(strArr2[index]);
if (cur_str1 < cur_str2) {
return -1;
} else if (cur_str1 > cur_str2) {
return 1;
}
index += 1;
} if (index < strArr1.length) {
for (int i = index; i < strArr1.length; i++) {
if (Integer.parseInt(strArr1[i]) > 0) {
return 1;
}
}
}
if (index < strArr2.length) {
for (int j = index; j < strArr2.length; j++) {
if (Integer.parseInt(strArr2[j]) > 0) {
return -1;
}
}
}
return 0;
}
}

Solution 2:

class Solution {
public int compareVersion(String version1, String version2) {
String[] strArr1 = version1.split("\\.");
String[] strArr2 = version2.split("\\.");
int len = Math.max(strArr1.length, strArr2.length);
for (int i = 0; i< len; i++) {
int cur_str1 = i >= strArr1.length ? 0 : Integer.parseInt(strArr1[i]);
int cur_str2 = i >= strArr2.length ? 0 : Integer.parseInt(strArr2[i]);
if (cur_str1 < cur_str2) {
return -1;
} else if (cur_str1 > cur_str2) {
return 1;
}
}
return 0;
}
}

[LC] 165. Compare Version Numbers的更多相关文章

  1. 【LeetCode】165. Compare Version Numbers 解题报告(Python)

    [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 ...

  2. 165. Compare Version Numbers - LeetCode

    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: p ...

  3. 【刷题-LeetCode】165 Compare Version Numbers

    Compare Version Numbers Compare two version numbers version1 and version2. If *version1* > *versi ...

  4. 165. Compare Version Numbers比较版本号的大小

    [抄题]: Compare two version numbers version1 and version2.If version1 > version2 return 1; if versi ...

  5. ✡ leetcode 165. Compare Version Numbers 比较两个字符串数字的大小 --------- java

    Compare two version numbers version1 and version2.If version1 > version2 return 1, if version1 &l ...

  6. Java for LeetCode 165 Compare Version Numbers

    Compare two version numbers version1 and version2.If version1 > version2 return 1, if version1 &l ...

  7. 【LeetCode】165 - Compare Version Numbers

    Compare two version numbers version1 and version2.If version1 > version2 return 1, if version1 &l ...

  8. Java [Leetcode 165]Compare Version Numbers

    题目描述: Compare two version numbers version1 and version2.If version1 > version2 return 1, if versi ...

  9. 165. Compare Version Numbers

    题目: Compare two version numbers version1 and version2.If version1 > version2 return 1, if version ...

随机推荐

  1. VNC/XRDP/XDMCP尝试

    (记得安装X Window System等 可参考链接https://www.linuxidc.com/Linux/2017-10/147646.htm) 看本文档时可以参考 https://blog ...

  2. h5-圆角的使用-案例安卓机器人

    1.圆角的使用 <!DOCTYPE html> <html lang="en"> <head> <meta charset="U ...

  3. CodeForces 1294B Collecting Packages(排序+贪心)

    http://codeforces.com/contest/1294/problem/B 大致题意: 一张图上有n个包裹,给出他们的坐标,一个机器人从(0,0)出发,只能向右(R)或向上(U),问能否 ...

  4. 干货 | CDN搭配OSS最佳实践 ——搭建动静态分离的应用架构

    一.传统架构及痛点 传统的网站产品应用架构,所有资源部署在应用服务器本地存储或挂载的数据存储区,对于动静态资源不作分离, 产品架构如下图所示: 该架构存在诸多问题: ● 系统性能会随着系统访问量的增长 ...

  5. 15)png图片旋转贴图

    1)基本代码展示  还是上一个那个总代码: #include<Windows.h> #include<gdiplus.h>//GDI+的头文件 using namespace ...

  6. 01 语言基础+高级:1-10 JDK8新特性_day12【函数式接口】

    day12[函数式接口] 主要内容自定义函数式接口函数式编程常用函数式接口 教学目标能够使用@FunctionalInterface注解能够自定义无参无返回函数式接口能够自定义有参有返回函数式接口能够 ...

  7. ccf201403-3 记录一个神tmwa了的代码 莫非我没看懂题。。。

    #include <string.h> #include<cstdio> #include<stdio.h> #include <iostream> # ...

  8. 线性齐次递推式快速求第n项 学习笔记

    定义 若数列 \(\{a_i\}\) 满足 \(a_n=\sum_{i=1}^kf_i \times a_{n-i}\) ,则该数列为 k 阶齐次线性递推数列 可以利用多项式的知识做到 \(O(k\l ...

  9. 四、NOSQL之Redis持久化缓存服务基础实战第三部

    1.NOSQL的理解 NOSQL是不仅仅是SQL,说的就是sql的补充,但是不能替代SQL. nosql库:memcached.memcachedb.redis 2.redis 简介 Redis是一个 ...

  10. tensorflow用法记录

    使用 embedding 变量 import tensorflow as tf import numpy as np sess = tf.InteractiveSession() M = list(' ...