leetcode253 - Meeting Rooms II - medium

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

 

Min heap. 先依据start time sort intervals, 之后依据end time开一个最小堆。遍历处理好的intervals，比较当前interval的start time和堆顶（end time), 如果end >= start，那就可以接着这个room继续用，pop堆顶（相当于清空这个房间给别人），否则别pop(相当于重新开一个房间)，每次都把当前interval的end time push进去。最后heap size就是需要的房间数。

c++里pq是max heap by default，min heap用struct greater即可。

sort custom objects最方便用lambda expression

// generic lambda, operator() is a template with two parameters
auto glambda = [](auto& a, auto& b) { return a < b; };

Detour一下。堆即一个完全二叉树，每个子树也满足一个堆数。如果父节点总是大于等于子节点就是最大堆，反之父节点总是小于等于子节点就是最小堆。插入/删除O(logn)

 

实现：Time O(nlogn) Space O(n)

class Solution {
public:
    int minMeetingRooms(vector<vector<int>>& intervals) {
        sort(intervals.begin(), intervals.end(), 
             [](const vector<int>& a, const vector<int>& b){return a[0] < b[0];});
        priority_queue<int, vector<int>, greater<int>> pq;
        for (auto interval : intervals){
            if (!pq.empty() && pq.top() <= interval[0]){
                pq.pop();
            }
            pq.push(interval[1]);
        }
        return pq.size();
    }
};