I come up this idea with the help of : https://discuss.leetcode.com/topic/30610/super-easy-java-solution-beats-98-8/2

However, I think my solution is O(n), cheaper than O(N logN)

```
public class Solution {
public int minMeetingRooms(Interval[] intervals) {
if(intervals.length == 0) return 0;
int maxtime = Integer.MIN_VALUE;
//Find max meeting time
for(int i = 0; i < intervals.length; i++){
maxtime = Math.max(maxtime, intervals[i].end);
}
int[] cont = new int[maxtime+1];
for(int i = 0; i < intervals.length; i++){
cont[intervals[i].start]++;
cont[intervals[i].end]--;
}
int res = 0;
for(int i = 1; i <= maxtime; i++){
cont[i] += cont[i-1];
res = Math.max(res, cont[i]);
}
return res;
}
}
```