[LeetCode] 252. Meeting Rooms 会议室

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.

给一个会议时间区间的数组，问一个人是否能参加所有的会议。就是求这些区间是否有交集，如果有就不能参加所有的会，没有交集就可以参加所有的会议。

解法：先把区间按开始时间排序，然后判断每一个区间的开始时间是否比前一个会议的结束时间早，如果有，就是时间重叠。

Java:

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public boolean canAttendMeetings(Interval[] intervals) {     Arrays.sort(intervals, new Comparator<Interval>(){         public int compare(Interval a, Interval b){             return a.start-b.start;         }     });        for(int i=0; i<intervals.length-1; i++){         if(intervals[i].end>intervals[i+1].start){             return false;         }     }        return true; }

Python：

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# Time:  O(nlogn) # Space: O(n) # # Definition for an interval. # class Interval: #     def __init__(self, s=0, e=0): #         self.start = s #         self.end = e   class Solution:     # @param {Interval[]} intervals     # @return {boolean}     def canAttendMeetings(self, intervals):         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

C++:

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class Solution { public:     bool canAttendMeetings(vector<Interval>& intervals) {         sort(intervals.begin(), intervals.end(), [](const Interval &a, const Interval &b){return a.start < b.start;});         for (int i = 1; i < intervals.size(); ++i) {             if (intervals[i].start < intervals[i - 1].end) {                 return false;             }         }         return true;     } };

 

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[LeetCode] 253. Meeting Rooms II 会议室 II

 

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