For example,
Given [[0, 30],[5, 10],[15, 20]],
return 2.
[分析]
初始思路,使用一个变量availableTime表示当前所有已安排会议的最早结束时间,对于每个待安排会议,比较其start 和 availableTime ,若start >= availableTime ,则不冲突,无需新增新会议室,否则需要增加新会议室。考虑case: [[1,5],[8,9],[8,9]] 就能理解该思路的bug。
由此可知仅维护一个availableTime变量不够,availableTime只能用一次。
修正bug,使用最小堆来维护所有已安排会议的结束时间,来一个新会议,仍然是比较其start 和 当前最早结束会议时间,若 start >= 最早结束时间,说明该会议可安排,就安排在最早结束那个会议的room,需要从最小堆中删除堆顶元素,将新会议的结束时间插入堆中,否则新增会议室。
思路2,参考
https://leetcode.com/discuss/50793/my-python-solution-with-explanation
原始注解:
# Very similar with what we do in real life. Whenever you want to start a meeting,
# you go and check if any empty room available (available > 0) and
# if so take one of them ( available -=1 ). Otherwise,
# you need to find a new room someplace else ( numRooms += 1 ).
# After you finish the meeting, the room becomes available again ( available += 1 ).
/** * 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; } * } */ public class Solution { Comparator<Interval> comparator = new Comparator<Interval>() { public int compare(Interval a, Interval b) { return a.start - b.start; } }; // Method -1: 初始思路,fail @ [[1,5],[8,9],[8,9]] public int minMeetingRooms(Interval[] intervals) { if (intervals == null || intervals.length == 0) return 0; Arrays.sort(intervals, comparator); int N = intervals.length; int rooms = 1; int availableTime = intervals[0].end; for (int i = 1; i < N; i++) { if (intervals[i].start < availableTime) { rooms++; availableTime = Math.min(availableTime, intervals[i].end); } } return rooms; } // Method 1 public int minMeetingRooms1(Interval[] intervals) { if (intervals == null || intervals.length == 0) return 0; Arrays.sort(intervals, comparator); int N = intervals.length; int rooms = 1; PriorityQueue<Integer> minHeap = new PriorityQueue<Integer>(); minHeap.offer(intervals[0].end); for (int i = 1; i < N; i++) { if (intervals[i].start < minHeap.peek()) { rooms++; } else { minHeap.poll(); } minHeap.offer(intervals[i].end); } return rooms; } // Method 2: https://leetcode.com/discuss/50793/my-python-solution-with-explanation public int minMeetingRooms(Interval[] intervals) { if (intervals == null || intervals.length == 0) return 0; int N = intervals.length; int[] starts = new int[N]; int[] ends = new int[N]; for (int i = 0; i < intervals.length; i++) { starts[i] = intervals[i].start; ends[i] = intervals[i].end; } Arrays.sort(starts); Arrays.sort(ends); int e = 0, rooms = 0, available = 0; for (int start : starts) { while (ends[e] <= start) { available++; e++; } if (available > 0) available--; else rooms++; } return rooms; } }