[LeetCode]Meeting Rooms

题目链接

只要按start time sort然后看所有interval是不是两两不相交即可。时间复杂度O(n * log n)，代码如下：

	/**
	* Definition for an interval.
	* struct Interval {
	* int start;
	* int end;
	* Interval() : start(0), end(0) {}
	* Interval(int s, int e) : start(s), end(e) {}
	* };
	*/
	class Solution {
	public:
	bool canAttendMeetings(vector<Interval>& intervals) {
	int len = intervals.size();
	auto comp = [](const Interval& lhs, const Interval& rhs)
	{
	return lhs.start < rhs.start;
	};
	sort(intervals.begin(), intervals.end(), comp);
	for(int i = 0; i < len - 1; ++i)
	{
	if(intervals[i].end > intervals[i + 1].start)return false;
	}
	return true;
	}
	};

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