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.原题地址 一道easy题,然而我zz一直没有反应过来。首先我们按照start time进行排序,要保证没有重叠的时间段,那么每一次的开始时间不能小于前一次的结束时间,即:一个会开完才能开下一个会。好吧,就这么简单。 12345678910111213141516171819202122232425 * 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 { // Define a compare function, sort intervals according to start time static bool (const Interval& itv1, const Interval& itv2) { return itv1.start < itv2.start; }public: bool canAttendMeetings(vector<Interval>& intervals) { sort(intervals.begin(), intervals.end(), comp); for (int i = 1; i < intervals.size(); i++) { if (intervals[i].start < intervals[i-1].end) { return false; } } return true; }}; 赞微海报分享
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