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Leetcode Meeting Rooms

时间:2015-10-06 06:57:22      阅读:218      评论:0      收藏:0      [点我收藏+]

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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.


解题思路:

先对start sort, O(nlgn), 然后比较Si+1 >= Ei, if true, continue compare, if false, return false. Total Complexity : O(nlgn)

问题: 注意Java 里的Arrays.sort(intervals, new Comparator<Interval>(){...} ); 使用方法


Java code:

/**
 * 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.length <= 1) {
            return true;
        }
        Arrays.sort(intervals, new Comparator<Interval>(){
            public int compare(Interval a, Interval b) {
                return a.start - b.start;
            }
        });
        for(int i = 1; i < intervals.length; i++) {
            if(intervals[i].start < intervals[i-1].end) {
                return false;
            }
        }
        return true;
    }
}

Reference:

1. https://leetcode.com/discuss/50912/ac-clean-java-solution

 

Leetcode Meeting Rooms

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原文地址:http://www.cnblogs.com/anne-vista/p/4856639.html

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