LeetCode 252. Meeting Rooms（会议室）

原题网址：https://leetcode.com/problems/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.

思路：按照开始时间排序，排好序之后，检查相邻的时间段是否重叠，方法是看start是否在另一个时间段的start和end之间（因为

start已经排好序，所以只检查这个条件就足够了）

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public boolean canAttendMeetings(Interval[] intervals) {
        Arrays.sort(intervals, new Comparator() {
           @Override
           public int compare(Interval i1, Interval i2) {
               return i1.start - i2.start;
           }
        });
        for(int i=0; i

方法二：利用start

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
public class Solution {
    public boolean canAttendMeetings(Interval[] intervals) {
        int[] starts = new int[intervals.length];
        int[] ends = new int[intervals.length];
        for(int i=0; i starts[i]) return false;
        }
        return true;
    }
}