252. Meeting Rooms

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252. Meeting Rooms

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5. Longest Palindromic Substring

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252. Meeting Rooms

Given an array of meeting time intervals where intervals[i] = [starti, endi],
determine if a person could attend all meetings.
想法：
將結束時間比較早的先做，所以排序是用 end time。
所以，如果剩下的的會議，其開始時間早於正在執行的會議結束時間，就回傳 false。
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class Solution {
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public:
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    bool canAttendMeetings(vector<vector<int>>& intervals) {
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        sort(intervals.begin(), intervals.end(), [](auto const& a, auto const& b) {
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            //sort each meeting by its end time
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            return a[1] < b[1];
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        });
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        for (int i = 1; i < intervals.size(); i++) {
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            //if a meeting ends after the next meeting starts you cannot attend both
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            if (intervals[i][0] < intervals[i - 1][1]){
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                return false;
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            }
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        }
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        return true;
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    }
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};
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