> For the complete documentation index, see [llms.txt](https://rachel2011.gitbook.io/leetcode_cc150/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://rachel2011.gitbook.io/leetcode_cc150/leetcode/meeting-rooms.md).

# Meeting Rooms

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]]`,\
return`false`.

这道题给了我们一堆会议的时间，问我们能不能同时参见所有的会议，这实际上就是求区间是否有交集的问题，我们可以先给所有区间排个序，用起始时间的先后来排，然后我们从第二个区间开始，如果开始时间早于前一个区间的结束时间，则说明会议时间有冲突，返回false，遍历完成后没有冲突，则返回true.

```
# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e

class Solution(object):
    def canAttendMeetings(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: bool
        """
        // Sort the intervals by start time
        intervals.sort(key = lambda x: x.start)
        for i in xrange(1, len(intervals)):
            if intervals[i].start < intervals[i-1].end:
                return False

        return True
```

**Complexity Analysis**

* Time complexity :O(nlog n) The time complexity is dominated by sorting. Once the array has been sorted, onlyO(n)time is taken to go through the array and determine if there is any overlap.
* Space complexity :O(1). Since no additional space is allocated.

## <https://leetcode.com/problems/meeting-rooms/#/solution>