的开始时间大于等于当前的结束时间, # 表示一个会议已经结束,可以释放一个房间 current_rooms -= 1 end_pointer += 1 # 更新需要的最大会议室数量 max_rooms = max(max_rooms, current_rooms) return max_rooms # 测试用例 intervals = [[0, 30], [5, 10], [15, 20]] print(minMeetingRooms(...
Given an array of meeting time intervals consisting of start and end times[[s1,e1],[s2,e2],...](si < ei), find the minimum number of conference rooms required. For example, Given[[0, 30],[5, 10],[15, 20]], return2. 252. Meeting Rooms的拓展,同样给一个开会的区间数组,返回最少...
1classSolution {2publicintminMeetingRooms(int[][] intervals) {3//corner case4if(intervals ==null|| intervals.length == 0) {5return0;6}78//normal case9Arrays.sort(intervals, (a, b) -> a[0] - b[0]);10//sort by ending time11PriorityQueue<int[]> heap =newPriorityQueue<>(intervals...
代码实现: importjava.util.*;publicclassMeetingRoomsII{publicstaticintminMeetingRooms(int[][]intervals){if(intervals.length==0){return0;}intn=intervals.length;// Step 1: Create two arrays to store the start and end timesint[]startTimes=newint[n];int[]endTimes=newint[n];for(inti=0;i<n...
class Solution { private: priority_queue<int, vector<int>, greater<int>> q; static bool cmp(const vector<int>& a, const vector<int>& b) { if (a[0] == b[0]) return a[1] < b[1]; return a[0] < b[0]; } public: int minMeetingRooms(vector<vector<int>>& intervals) { /...
【LeetCode】253. Meeting Rooms II 解题报告(C++) id: fuxuemingzhu http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 排序+堆 日期 题目地址:https://leetcode-cn.com/problems/meeting-rooms/ 题目描述 Given an array of meeting time intervals consisting of start and end times[[s1,e1],[...
Meeting Rooms II Given an array of meeting time intervals consisting of start and end times[[s1,e1],[s2,e2],...] (si < ei), find the minimum number of conference rooms required. For example, Given[[0, 30],[5, 10],[15, 20]], return 2. ...
//#253Description: Meeting Rooms II | LeetCode OJ 解法1:看代码,不解释。 // Solution 1: Let the code explain itself. 代码1 //Code 1 254 Factor Combinations // #254 因式组合 描述:给定正整数n,返回所有乘积等于n的不同组合,除了n = n。为避免重复,所有乘数单调递增。
输出:false示例2:输入: 2. 解题 按开始时间排序后,依次检查相邻前一个的结束和后一个的开始时间是否重叠 代码语言:javascript 代码运行次数:0 运行 AI代码解释 classSolution{//C++public:boolcanAttendMeetings(vector<vector<int>>&intervals){sort(intervals.begin(),intervals.end(),[&](auto a,auto b){ret...
253Meeting Rooms II☢ 252Meeting Rooms☢ 251Flatten 2D Vector☢ 250Count Univalue Subtrees☢ 249Group Shifted Strings☢ 248Strobogrammatic Number III☢ 247Strobogrammatic Number II☢ 246Strobogrammatic Number☢ 245Shortest Word Distance III☢ ...