leetcode-252. Meeting Rooms

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。

/**
 * 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 comp(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;
    }
};