Github 同步地址: https://github.com/grandyang/leetcode/issues/252 类似题目: Merge Intervals Meeting Rooms II 参考资料: https://leetcode.com/problems/meeting-rooms/ https://leetcode.com/problems/meeting-rooms/discuss/67782/C%2B%2B-sort https://leetcode.com/problems/meeting-rooms/discuss/67786/...
class Solution { public int mostBooked(int n, int[][] meetings) { // sort meetings according to start time, simulate real meeting arrangement Arrays.sort(meetings, (a, b) -> Integer.compare(a[0], b[0])); // track available meeting rooms TreeSet<Integer> availableRooms = new TreeSe...
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. 贪心法 复杂度 时间O(NlogN) 空间 O(...
// Step 3: Scan the events and maintain the number of meeting rooms in use. int maxRooms = 0; int roomsInUse = 0; for (int[] event : events) { // Update the number of rooms in use roomsInUse += event[1]; // Update the maximum number of rooms needed maxRooms = Math.max(...
heappush(free_rooms, intervals[0][1]) # 从第二个会议开始遍历 for i in intervals[1:]: # 如果当前会议的开始时间大于等于最小堆中的最早结束时间 # 说明这个会议室可以被重复使用 # 因此我们可以移除堆顶元素(最早结束的会议室) if i[0] >= free_rooms[0]: heapq.heappop(free_rooms) # 将当前...
则可以复用房间if(startTimes[i]>=endTimes[endIndex]){// Reuse the room: move the endIndex to the next meetingendIndex++;}else{// If no room is available, we need a new oneroomCount++;}}// The room count will be the number of rooms we needreturnroomCount+1;// We need at least...
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],[s2,e2],...] (si < ei), find the minimum number of conference rooms required. ...
meeting room II : greedy algorithm and compare function question: 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 exampl......
0253-Meeting-Rooms-II/cpp-0253 CMakeLists.txt main.cpp main2.cpp main3.cpp main4.cpp 0254-Factor-Combinations 0257-Binary-Tree-Paths 0259-3Sum-Smaller 0268-Missing-Number 0279-Perfect-Squares 0282-Expression-Add-Operators 0283-Move-Zeroes 0286-Walls-and-Gates 0287-Find-t...
The point(0,2)is an ideal meeting point, as the total travel distance of2+2+2=6is minimal. So return 6. 横纵分离 复杂度 时间O(NM) 空间 O(NM) 思路 为了保证总长度最小,我们只要保证每条路径尽量不要重复就行了,比如1->2->3<-4这种一维的情况,如果起点是1,2和4,那2->3和1->2->3这...