[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;
}
}
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