253. Meeting Rooms II

问题描述：

Given an array of meeting time intervals consisting of start and end times [[s1,e1],[s2,e2],...] (si < ei), find the minimum number of conference rooms required.

Example 1:

Input: [[0, 30],[5, 10],[15, 20]]
Output: 2

Example 2:

Input: [[7,10],[2,4]]
Output: 1

解题思路：

要最少的最多的房间，我们只要满足时段内重复最多的不同会议就好。

可以将会议开始和结束都存到最小堆里，用pair存储，第二个值用于标识是开始还是结束。

我从最小堆中取出堆顶元素（即最小值）

若其为起点，则rooms自增一；若其为重点，首先将当前rooms的值与ret相比，取最大的那个，然后rooms自减一。

还有一个很神奇的解法

代码：

用最小堆来解：

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    int minMeetingRooms(vector<Interval>& intervals) {
        priority_queue<pair<int,int>, vector<pair<int,int>>, greater<> > q;
        int ret = 0;
        for(Interval i : intervals){
            q.push({i.start, 1}); //represent start
            q.push({i.end, -1}); //represent end
        }
        
        int rooms = 0;
        while(!q.empty()){
            auto p = q.top();
            q.pop();
            if(p.second == 1){
                //represent start
                rooms++;
            }else{
                //represent end
                ret = max(ret, rooms);
                rooms--;
            }
        }
        
        
        return ret;
    }
};

很神奇的解法：

/**
 * Definition for an interval.
 * struct Interval {
 *     int start;
 *     int end;
 *     Interval() : start(0), end(0) {}
 *     Interval(int s, int e) : start(s), end(e) {}
 * };
 */
class Solution {
public:
    int minMeetingRooms(vector<Interval>& intervals) {
        vector<int> starts, ends;
        int res = 0, endpos = 0;
        for (auto a : intervals) {
            starts.push_back(a.start);
            ends.push_back(a.end);
        }
        sort(starts.begin(), starts.end());
        sort(ends.begin(), ends.end());
        for (int i = 0; i < intervals.size(); ++i) {
            if (starts[i] < ends[endpos]) ++res;
            else ++endpos;
        }
        return res;
    }
};