[LeetCode] 252. Meeting Rooms 会议室

 


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 1:

Input: [[0,30],[5,10],[15,20]]
Output: false

Example 2:

Input: [[7,10],[2,4]]
Output: true

NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.

 

这道题给了我们一堆会议的时间,问能不能同时参见所有的会议,这实际上就是求区间是否有交集的问题,那么最简单暴力的方法就是每两个区间比较一下,看是否有 overlap,有的话直接返回 false 就行了。比较两个区间a和b是否有 overlap,可以检测两种情况,如果a的起始位置大于等于b的起始位置,且此时a的起始位置小于b的结束位置,则一定有 overlap,另一种情况是a和b互换个位置,如果b的起始位置大于等于a的起始位置,且此时b的起始位置小于a的结束位置,那么一定有 overlap,参见代码如下:

 

解法一:

class Solution {
public:
    bool canAttendMeetings(vector<vector<int>>& intervals) {
        for (int i = 0; i < intervals.size(); ++i) {
            for (int j = i + 1; j < intervals.size(); ++j) {
                if ((intervals[i][0] >= intervals[j][0] && intervals[i][0] < intervals[j][1]) || (intervals[j][0] >= intervals[i][0] && intervals[j][0] < intervals[i][1])) return false;
            }
        }
        return true;
    }
};

 

我们可以先给所有区间排个序,用起始时间的先后来排,然后从第二个区间开始,如果开始时间早于前一个区间的结束时间,则说明会议时间有冲突,返回 false,遍历完成后没有冲突,则返回 true,参见代码如下:

 

解法二:

class Solution {
public:
    bool canAttendMeetings(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end(), [](const vector<int>& a, const vector<int>& b){return a[0] < b[0];});
        for (int i = 1; i < intervals.size(); ++i) {
            if (intervals[i][0] < intervals[i - 1][1]) {
                return false;
            }
        }
        return true;
    }
};

 

Github 同步地址:

https://github.com/grandyang/leetcode/issues/252

 

类似题目:

Merge Intervals

Meeting Rooms II

 

参考资料:

https://leetcode.com/problems/meeting-rooms/

https://leetcode.com/problems/meeting-rooms/discuss/67782/C%2B%2B-sort

https://leetcode.com/problems/meeting-rooms/discuss/67786/AC-clean-Java-solution

 

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posted @ 2016-03-04 07:13 Grandyang 阅读(...) 评论(...) 编辑 收藏