[leetcode]253. Meeting Rooms II 会议室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.

Example 1:

Input: [[0, 30],[5, 10],[15, 20]]
Output: 2

Example 2:

Input: [[7,10],[2,4]]
Output: 1

思路

| 0 ------------------------- 30 |

|5-----10|

|15---20|

1.  if starts[i] > ends[endItr],  which means  current meeting starts right after, we can continue to use previous meeting room. All we need to do is update ends pointer

2. otherwies, starts[i] < ends[endItr],  which means current meeting starts when previous meeting has not ended. Then we need a new room for current meeting.

代码

 class Solution {
      public int minMeetingRooms(Interval[] intervals) {
         int[] starts = new int[intervals.length];
         int[] ends = new int[intervals.length];
         for(int i = 0; i< intervals.length; i++) {
             starts[i] = intervals[i].start;
             ends[i] = intervals[i].end;
         }
         Arrays.sort(starts);
         Arrays.sort(ends);
         int rooms = 0;
         int endsItr = 0;
         for(int i= 0; i< starts.length; i++) {
             if(starts[i]<ends[endsItr])
                 rooms++;
             else
                 endsItr++;
         }
         return rooms;
     }
 }