LeetCode 题解(255) : Meeting Rooms II

题目：

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

For example,
Given [[0, 30],[5, 10],[15, 20]],
return 2.

题解：

两种做法。都需要先根据start排序原数组。然后要么用最小堆，要么对每一个start，遍历end，比较end是否小于等于start。

C++的priority_queue是最大堆，Java的PriorityQueue和Python的heapq是最小堆。面试的时候可以需求选择适当的语言。

C++版：

/**
 * 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& intervals) {
        vector s(intervals.size());
        vector e(intervals.size());
        for(int i = 0; i < intervals.size(); i++) {
            s[i] = intervals[i].start;
            e[i] = intervals[i].end;
        }
        sort(s.begin(), s.end());
        sort(e.begin(), e.end());
        int eIndex = 0, room = 0, available = 0;
        for(auto i : s) {
            while(e[eIndex] <= i) {
                available++;
                eIndex++;
            }
            if(available > 0)
                available--;
            else
                room++;
        }
        return room;
    }
};

Java版：

/**
 * Definition for an interval.
 * public class Interval {
 *     int start;
 *     int end;
 *     Interval() { start = 0; end = 0; }
 *     Interval(int s, int e) { start = s; end = e; }
 * }
 */
import java.util.Arrays; 
 
public class Solution {
    public int minMeetingRooms(Interval[] intervals) {
        Arrays.sort(intervals, new IntervalComparator());
        PriorityQueue minHeap = new PriorityQueue();
        int rooms = 0;
        for(int i = 0; i < intervals.length; i++) {
            if(minHeap.size() == 0) {
                minHeap.add(intervals[i].end);
                rooms++;
                continue;
            }
            if(minHeap.peek() <= intervals[i].start) {
                minHeap.poll();
                minHeap.add(intervals[i].end);
            } else {
                minHeap.add(intervals[i].end);
                rooms++;
            }
        }
        return rooms;
    }
}

class IntervalComparator implements Comparator {
    public int compare(Interval a, Interval b) {
        return a.start - b.start;
    }
}

Python版：

# Definition for an interval.
# class Interval(object):
#     def __init__(self, s=0, e=0):
#         self.start = s
#         self.end = e
import heapq

class Solution(object):
    def minMeetingRooms(self, intervals):
        """
        :type intervals: List[Interval]
        :rtype: int
        """
        heap, num = [], 0
        heapq.heapify(heap)
        intervals.sort(lambda a, b : a.start - b.start)
        for i in range(len(intervals)):
            if len(heap) == 0:
                heapq.heappush(heap, intervals[i].end)
                num += 1
                continue
            if heap[0] <= intervals[i].start:
                heapq.heappop(heap)
                heapq.heappush(heap, intervals[i].end)
            else:
                heapq.heappush(heap, intervals[i].end)
                num +=1
                
        return num