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/AC-clean-Java-solution
public int minMeetingRooms(Interval[] intervals) { if(intervals == null || intervals.length == 0) return 0; Arrays.sort(intervals, new Comparator<Interval>(){ public int compare(Interval i1, Interval i2){ return i1.start - i2.start; } }); // 用堆来管理房间的结束时间 PriorityQueue<...
[LeetCode] Meeting Rooms Problem Description: Given an array of meeting time intervals consisting of start and end times[[s1,e1],[s2,e2],...](si < ei), determine if a person could attend all meetings. For example, Given[[0, 30],[5, 10],[15, 20]], returnfalse. The idea is p...
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(...
classSolution{//C++public:boolcanAttendMeetings(vector<vector<int>>&intervals){sort(intervals.begin(),intervals.end(),[&](auto a,auto b){returna[0]<b[0];});for(int i=1;i<intervals.size();++i){if(intervals[i-1][1]>intervals[i][0])returnfalse;}returntrue;}}; ...
解法1:删呗。 // Solution 1: Do it. 代码1 //Code 1 204 Count Primes // #204 数质数 描述:统计小于n的质数个数。 //#204Description: Count Primes | LeetCode OJ 解法1:筛法。 // Solution 1: Sieve of Eratosthenes. 代码1 //Code 1 ...
这道题是之前那道Meeting Rooms的拓展,那道题只让我们是否能参加所有的会,也就是看会议之间有没有时间冲突,而这道题让我们求最少需要安排几个会议室,有时间冲突的肯定需要安排在不同的会议室。这道题有好几种解法,我们先来看使用TreeMap来做的,我们遍历时间区间,对于起始时间,映射值自增1,对于结束时间,映射值...
int minMeetingRooms(vector<vector<int>>& a) { int n = a.size(); vector<int> s(n,0); vector<int> e(n,0); for(int i=0;i<a.size();++i){ s[i] = a[i][0]; e[i] = a[i][1]; } sort(s.begin(),s.end()); ...
{ 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) { // 1.按开始...
Meeting Rooms Swift Easy O(nlogn) O(1) Meeting Rooms II Swift Medium O(nlogn) O(n) Merge Intervals Swift Hard O(nlogn) O(n) Alien Dictionary Swift Hard O(nm) O(nm) Kth Largest Element in an Array Swift Medium O(nlogn) O(n) Array Partition I Swift Easy O(nlogn) O(n) Inse...