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  • 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.

    Implement a Comparator<Interval>

    Syntax: don't forget the public sign when defining a function

    /**
     * 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) {
            if (intervals==null || intervals.length==0 || intervals.length==1) return true;
            Comparator<Interval> comp = new Comparator<Interval>() {
                public int compare(Interval i1, Interval i2) {
                    return (i1.start==i2.start)? i1.end-i2.end : i1.start-i2.start;
                }
            };
            
            Arrays.sort(intervals, comp);
            Interval pre = intervals[0];
            for (int i=1; i<intervals.length; i++) {
                Interval cur = intervals[i];
                if (cur.start < pre.end) return false;
                pre = cur;
            }
            return true;
        }
    }
    

      

    public static <T> void sort(T[] a,
                                Comparator<? super T> c)
    根据指定比较器产生的顺序对指定对象数组进行排序。
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  • 原文地址:https://www.cnblogs.com/apanda009/p/7354220.html
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