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.
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根據start point sort, 再traverse整個vector, 如果後面一個meeting開始的時間比之前一個meeting結束的時間早(即有重疊)則return false。
記得有好幾道題要寫customize compare function的。
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;
}
};