- 综合排序
- 最热优先
- 最新优先
- 来自专栏福大大架构师每日一题
2022-01-01：给定int meetings，比如 { {66, 70} 0号会议
9, 50}, {13, 42}} ret := maxScore1(meetings) fmt.Println(ret) ret = maxScore2(meetings) fmt.Println(ret) } func maxScore1(meetings [][]int) int { //Arrays.sort(meetings, (a, b) -> a[0] - b[0]); sort.Slice(meetings, func(i, j int) bool { return meetings[i][0] < meetings[j][ 0] }) //int[][] path = new int[meetings.length][]; path0 := make([][]int, len(meetings)) [][]int) int { sort.Slice(meetings, func(i, j int) bool { return meetings[i][0] < meetings
16210编辑于 2022-01-01 - 来自专栏js笔记
会议室, 输入是一个数组, 所有会议的开始和结束时间. 输出一共需要多少个会议室
=> { // 会议按开始时间排序 if (meetings == null || meetings.length == 0) return 0; meetings.sort ; i++) { if (meetingEnd <= meetings[i][0]) { meetingEnd = Math.min(meetingEnd, meetings if (meetings == null || meetings.length == 0) return 0; var start = []; var end = []; for (var i = 0; i < meetings.length; i++) { start[i] = meetings[i][0]; end[i] = meetings i = 0; i < meetings.length; i++) { start[i] = meetings[i][0]; end[i] = meetings[i][1];
90420编辑于 2022-10-25 - 来自专栏福大大架构师每日一题
2025-01-06：无需开会的工作日。用go语言，给定一个正整数 days，表示员工可以工作的总天数（从第 1 天开始）。 同
同时还有一个二维数组 meetings，其长度为 n，其中每个元素 meetings[i] = [start_i, end_i] 表示第 i 次会议的开始和结束天数（包含这两天）。 1 <= meetings.length <= 100000。 meetings[i].length == 2。 1 <= meetings[i][0] <= meetings[i][1] <= days。 输入：days = 10, meetings = [[5,7],[1,3],[9,10]]。 输出：2。 [][2], int meetingCount){ // 排序会议 qsort(meetings, meetingCount,sizeof(meetings[0]), compare); int ) { // 按照左端点从小到大排序 std::sort(meetings.begin(), meetings.end()); int start =1, end =0;// 当前合并区间的左右端点
32520编辑于 2025-01-07 - 来自专栏福大大架构师每日一题
2022-06-09：每个会议给定开始和结束时间，后面的会议如果跟前面的会议有任何冲突，完全取消冲突的、之前的会议，安排当前的。
= random_meeting(len, t); let mut meetings2 = copy_meetings(&mut meetings1); let mut ans1 = arrange1(&mut meetings1); let mut ans2 = arrange2(&mut meetings2); if ! []; let mut i = meetings.len() as i32 - 1; while i >= 0 { let cur = &meetings[i as usize rank0[i as usize] = meetings[i as usize][0]; // 会议开头点 rank0[(i + n) as usize] = meetings[i as ; } // 为了测试 fn copy_meetings(meetings: &mut Vec<Vec<i32>>) -> Vec<Vec<i32>> { return meetings.clone
39620编辑于 2023-06-08 - 来自专栏福大大架构师每日一题
2022-05-10：在字节跳动，大家都使用飞书的日历功能进行会议室的预订，遇到会议高峰时期， 会议室就可能不够用，现在请你实现一个算法，判断预订会议时是否有空的会议室可用。
[9, 199, 320], ]; let answer = reserve_meetings(2, &mut meetings); println! let mut re_meetings: Vec<Vec<isize>> = vec![]; for i in 0..n { re_meetings.push(vec! max: isize = 0; for i in 0..n { re_meetings[i as usize][0] = i; re_meetings[i as usize][1] = meetings[i as usize][0]; re_meetings[i as usize][2] = rank(&mut ranks, meetings[ st.query_max(re_meetings[i as usize][2], re_meetings[i as usize][3]) < m { ans[re_meetings
46950编辑于 2022-06-04 - 来自专栏福大大架构师每日一题
2022-06-09：每个会议给定开始和结束时间， 后面的会议如果跟前面的会议有任何冲突，完全取消冲突的、之前的会议，安排当前的。 给定一个会议数组，返回安排的
= random_meeting(len, t); let mut meetings2 = copy_meetings(&mut meetings1); let mut ans1 = arrange1(&mut meetings1); let mut ans2 = arrange2(&mut meetings2); if ! []; let mut i = meetings.len() as i32 - 1; while i >= 0 { let cur = &meetings[i as usize rank0[i as usize] = meetings[i as usize][0]; // 会议开头点 rank0[(i + n) as usize] = meetings[i as ;}// 为了测试fn copy_meetings(meetings: &mut Vec<Vec<i32>>) -> Vec<Vec<i32>> { return meetings.clone()
68230编辑于 2022-06-09 - 来自专栏福大大架构师每日一题
2022-03-16：给你一个整数 n ，表示有 n 个专家从 0 到 n -
另外给一个下标从 0 开始的二维整数数组 meetings ， 其中 meetingsi = xi, yi, timei 表示专家 xi 和专家 yi 在时间 timei 要开一场会。 meetings, func(i, j int) bool { a := meetings[i] b := meetings[j] return a[2] < b[2] }) // [1,7,1 = meetings[0][1] size := 2 for i := 1; i < m; i++ { // i 2 if meetings[i][2] ! = meetings[i-1][2] { share(help, size, uf) help[0] = meetings[i][0] help[1] = meetings[i][1] size = 2 } else { help[size] = meetings[i][0] size++ help[size] = meetings[i][1] size
50420编辑于 2022-03-16 - 来自专栏CSDN博客专家-小蓝枣的博客
Python 算法高级篇：贪心算法的原理与应用
3.1 会议室安排问题 def max_meetings(meetings): if not meetings: return [] # 按结束时间升序排序 meetings.sort (key=lambda x: x[1]) result = [meetings[0]] prev_end = meetings[0][1] for meeting in meetings[1:]: start, end = meeting if start >= prev_end: result.append(meeting ) prev_end = end return result meetings = [(1, 3), (2, 4), (3, 5), (5, 7), (6, 8)] selected_meetings = max_meetings(meetings) print("Selected Meetings:") for meeting in selected_meetings
78930编辑于 2023-10-26 - 来自专栏福大大架构师每日一题
2022-05-10：在字节跳动，大家都使用飞书的日历功能进行会议室的预订，遇到会议高峰时期， 会议室就可能不够用，现在请你实现一个算法，判断预订会议时是否有空
[9, 199, 320], ]; let answer = reserve_meetings(2, &mut meetings); println! let mut re_meetings: Vec<Vec<isize>> = vec![]; for i in 0..n { re_meetings.push(vec! max: isize = 0; for i in 0..n { re_meetings[i as usize][0] = i; re_meetings[i as usize][1] = meetings[i as usize][0]; re_meetings[i as usize][2] = rank(&mut ranks, meetings[ st.query_max(re_meetings[i as usize][2], re_meetings[i as usize][3]) < m { ans[re_meetings
68610编辑于 2022-05-10 - 来自专栏Michael阿明学习之路
LeetCode 2092. 找出知晓秘密的所有专家（并查集）
另外给你一个下标从 0 开始的二维整数数组 meetings ，其中 meetings[i] = [xi, yi, timei] 表示专家 xi 和专家 yi 在时间 timei 要开一场会。 示例 1： 输入：n = 6, meetings = [[1,2,5],[2,3,8],[1,5,10]], firstPerson = 1 输出：[0,1,2,3,5] 解释： 时间 0 ，专家 0 提示： 2 <= n <= 10^5 1 <= meetings.length <= 10^5 meetings[i].length == 3 0 <= xi, yi <= n - 1 xi ! known[0] = known[firstPerson] = true; sort(meetings.begin(), meetings.end(),[&](auto& a, auto int a = meetings[j][0], b = meetings[j][1]; int f1 = uset.find(a); int
53640编辑于 2022-01-07 - 来自专栏我的充电站
LeetCode笔记：Weekly Contest 269
else: self.dsu[x] = y class Solution: def findAllPeople(self, n: int, meetings : List[List[int]], firstPerson: int) -> List[int]: known = {firstPerson, 0} meetings = sorted(meetings, key=lambda x: x[2]) i, m = 0, len(meetings) while i < m: j = i dsu = DSU(firstPerson) while j < m and meetings[j][2] == meetings [i][2]: dsu.union(meetings[j][0], meetings[j][1], known) j += 1
30920发布于 2021-12-01 - 来自专栏labuladong的算法专栏
时间调度问题的千层套路
函数签名如下： // 返回需要申请的会议室数量 int minMeetingRooms(int[][] meetings); 比如给你输入meetings = [[0,30],[5,10],[15,20 举例来说，如果输入meetings = [[0,30],[5,10],[15,20]]，那么我们就给数组中[0,30],[5,10],[15,20]这几个索引区间分别加一，最后遍历数组，求个最大值就行了 首先，对区间进行投影，就相当于对每个区间的起点和终点分别进行排序： int minMeetingRooms(int[][] meetings) { int n = meetings.length ) { int n = meetings.length; int[] begin = new int[n]; int[] end = new int[n]; for(int i = 0; i < n; i++) { begin[i] = meetings[i][0]; end[i] = meetings[i][1]; } Arrays.sort
1.4K20发布于 2021-09-23 - 来自专栏机器学习/数据可视化
SQL进阶-9-谓词exists使用
生成全部的会议的集合使用交叉联结： select distinct M1.meeting, M2.person from Meetings M1 cross join Meeting M2; -- 使用存在量化求解： select distinct M1.meeting, M2.person from Meetings M1 cross join Meetings M2 where not exists ( select * from Meetings M3 where M1.meeting = M3.meeting -- 不存在M3与M1和M2相同的数据：即用全部集合减去现有的表中的数据 and M2.person = M3.person ); 使用差集求解： select distinct M1.meeting, M2.person from Meetings M1 cross join Meetings M2 except -- 差集排除 not exists具备了差集的功能 select meeting, person from Meetings; 笔记：肯定=双重否定
1.4K20发布于 2021-03-01 - 来自专栏Web行业观察
【英文】Sponsorship Levels & Benefits
Benefits One of the primary membership benefits of TAPA Americas is the free access to our bi-annual “T” meetings Support for our meetings has traditionally come from TAPA members who provide meeting sites and/or financial toward facility, event materials, and catering Bronze Sponsor: Sponsorship of shuttle transportation to meetings
1K60发布于 2019-12-03 - 来自专栏专注数据中心高性能网络技术研发
[Repost]The care and maintenance of your adviser
Meetings A comment we often hear at our workshops is, “My adviser is lovely but he/she is just so busy Regardless, you need to organize meetings where you can get real face time and talk about your thesis We mean regularly scheduled meetings focusing on your thesis. You can't assume that your adviser hosts productive meetings or can intuit what you need to know.
870110发布于 2018-03-08 - 来自专栏SAMshare
[009] DS的Code Review该怎么做？这15条建议可以了解一下
This past year I have hosted weekly code review meetings where at least three to five developers and As I began to host these meetings, I have learned some valuable lessons on why code reviews are essential Exposure to New Ideas and Technologies The main reason I have found code review meetings beneficial on These informal code reviews are meetings in which a developer or data scientist invites one or two people These meetings happen well before the formal code review sessions and are useful when making decisions
55420发布于 2020-12-02 - 来自专栏YoungGy
R语言包_lubridate
"2013-01-01 UTC" 关于时间取整，取模的其他计算 meeting <- ymd_hms("2011-07-01 09:00:00", tz = "Pacific/Auckland") meetings <- interval(ymd(20110720, tz = "Pacific/Auckland"), ymd(20110831, tz = "Pacific/Auckland")) meeting meetings jsm meetings %within% jsm arrive <- ymd_hms("2011-06-04 12:00:00", tz = "Pacific/Auckland") leave <
1.5K30发布于 2019-05-26 - 来自专栏程序员成长充电站
10倍工程师和1倍工程师，10x engineer and 1x engineer
Shekhar Kirani在推特上说创业者要找到能以一扛十的工程师，即10倍工程师，这样创业成功的概率会大大提高，然后这哥们还列出了10倍工程师的特点，原文如下： 10x engineers hate meetings They attend meetings because the manager has called for a "Staff meeting" to discuss the features and They move out because you make their life miserable with the process, meetings, training, and other non-value-added Is a team player that goes to the same meetings their co-workers are required to go to.
93810发布于 2019-09-27 - 来自专栏音视频技术
Webex 如何在在线会议领域保持优势？
文/ Zeus Kerravala 译/ 元宝 原文 https://www.nojitter.com/video-collaboration-av/webex-meetings-or-zoom-peering-performance Ravichandran解释说，Webex有许多算法可以让Webex Meetings从网络拥塞中快速恢复。 其性能难以通过纸面数据量化（与数据中心的数量相比），因此我要求思科在相关领域投入资金研发技术，并测试比较Webex Meetings和Zoom之间的性能差异。 正如你期望的那样，这两项产品在720p的清晰度条件下都能提供高质量的在线视频会议使用体验，虽然Zoom Meetings的帧速率更高，但其消耗的带宽是Webex Meetings的两倍，如下图所示。
61820发布于 2021-09-01 - 来自专栏不能显示专栏创建者
FTC的Zoom Deal表示对安全执法的承诺
the cryptographic keys that could actually allow the company to access the content of its customers' meetings , and secured its teleconference meetings, in part, with a lower level of encryption than promised, FTC According to the FTC's complaint, Zoom also misled some users who wanted to store recorded meetings on the company's cloud storage by falsely claiming that those meetings were encrypted immediately after
80100发布于 2020-12-29
腾讯云开发者
Copyright © 2013 - 2026 Tencent Cloud. All Rights Reserved. 腾讯云 版权所有
深圳市腾讯计算机系统有限公司 ICP备案/许可证号：粤B2-20090059 
粤公网安备44030502008569号
腾讯云计算（北京）有限责任公司 京ICP证150476号 | 京ICP备11018762号