LintCode 920: Meeting Rooms

最新推荐文章于 2020-09-13 00:09:51 发布

原创最新推荐文章于 2020-09-13 00:09:51 发布 · 380 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#LintCode

algorithm-design 专栏收录该内容

817 篇文章

订阅专栏

博客围绕会议时间冲突判断问题展开，给定会议时间区间数组，需判断能否参加所有会议。通过示例展示不同输入的输出结果，如输入[(0,30),(5,10),(15,20)]输出为false，输入[(5,8),(9,15)]输出为true，并给出先排序再看是否重合的解法。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Meeting Rooms
中文English
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.

Example
Example1

Input: intervals = [(0,30),(5,10),(15,20)]
Output: false
Explanation:
(0,30), (5,10) and (0,30),(15,20) will conflict
Example2

Input: intervals = [(5,8),(9,15)]
Output: true
Explanation:
Two times will not conflict

解法1：
先排序，然后看是否有重合。
代码如下：

/**
 * Definition of Interval:
 * classs Interval {
 *     int start, end;
 *     Interval(int start, int end) {
 *         this->start = start;
 *         this->end = end;
 *     }
 * }
 */

struct cmp {
    bool operator() (Interval & a, Interval & b) {
        return a.start < b.start;
    }
} compare;

class Solution {
public:
    /**
     * @param intervals: an array of meeting time intervals
     * @return: if a person could attend all meetings
     */
    bool canAttendMeetings(vector<Interval> &intervals) {
        int n = intervals.size();
        sort(intervals.begin(), intervals.end(), compare);
        
        for (int i = 0; i < n - 1; ++i) {
            if (intervals[i].end > intervals[i + 1].start) return false;
        }
        
        return true;
    }
};