[Leetcode] Meeting Rooms

The idea was pretty simple:first we sort the in the ascending order of and then intervals start we check for the overlapping of E Ach pair of neighboring intervals. If they do, then return false ; After we finish all the checks and has not returned false , just return true .

Sorting takes time O(nlogn) and the overlapping checks take time O(n) , so this idea is time of total O(nlogn) .

The code is as follows.

1 classSolution {2  Public:3     BOOLCanattendmeetings (vector<interval>&intervals) {4 sort (Intervals.begin (), intervals.end (), compare);5         intn =intervals.size ();6          for(inti =0; I < n-1; i++)7             if(overlap (Intervals[i], intervals[i +1]))8                 return false;9         return true;Ten     } One Private: A     Static BOOLCompare (interval& Interval1, interval&Interval2) { -         returnInterval1.start <Interval2.start; -     } the     BOOLOverlap (interval& interval1, interval&Interval2) { -         returnInterval1.end >Interval2.start; -     }  -};

[Leetcode] Meeting Rooms