编程练习9 会议室占用时间段

这个主要是对边界条件的处理需要注意，比如只有一个会议，两个会议以及重叠与不重叠各种情况。

#include <iostream>
#include <vector>
#include <algorithm>

using namespace std;

struct MeetingTime {
    int start;
    int end;
};

bool compare(const MeetingTime &a, const MeetingTime &b) {
    return a.start < b.start;
}

int main() {
	int meeting_num = 0;
	cin >> meeting_num;
	vector<MeetingTime> meetings;
    for (int i = 0; i < meeting_num; i++) 
	{
		MeetingTime a;
		cin >> a.start >> a.end;
		meetings.push_back(a);
    }
    
    // 假设已经通过某种方式填充了meetings向量
    // 例如：meetings = {{1, 4}, {2, 5}, {7, 9}, {14, 18}};
    
    // 排序
    sort(meetings.begin(), meetings.end(), compare);
    
    // 合并时间段
    vector<MeetingTime> merged;
	int meetings_num = meetings.size();
    for (int i = 0; i < meetings_num; ++i) {
		//如果只有一个会议 直接输出
		if(meetings.size() == 1)
		{
			merged.push_back(meetings[i]);
		}
		//如果当前会议时间和后面一个会议重叠 则需要将两次会议合并 直接将合并后的会议替换下一个会议时间
        else if (meetings[i].end >= meetings[i + 1].start && i<meetings_num-1) 
		{
            MeetingTime curr = meetings[i];
            if (meetings[i].end < meetings[i + 1].end) {
                curr.end = meetings[i + 1].end;
            }
			
			meetings[i + 1].start = curr.start;
			meetings[i + 1].end = curr.end;		
        }
		else
		{
			//如果是倒数第二个会议 需要单独处理
			if(i == meetings.size()-2)
			{
				//如果与最后一个会议不重叠 则将当前会议输出
				if(meetings[i].end < meetings[i + 1].start)
				{
					merged.push_back(meetings[i]);
					//merged.push_back(meetings[i+1]);
				}
				else if(meetings[i].end >= meetings[i + 1].start)
				{
					 MeetingTime curr = meetings[i];
					if (meetings[i].end < meetings[i + 1].end) {
						curr.end = meetings[i + 1].end;
					}
					merged.push_back(curr);
				}

			}
			else
			{
				merged.push_back(meetings[i]);
			}
		}
		
    }
    
    // 输出结果
    for (const auto &m : merged) {
        cout << m.start << " " << m.end << endl;
    }
    
    return 0;
}