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原题链接在这里：https://leetcode.com/problems/meeting-rooms-ii/
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.
For example,
Given [[0, 30],[5, 10],[15, 20]],
return 2.
与Meeting Rooms类似。
排序后从前向后扫描。同时维护一个min heap, 把每个interval 的end 放到min heap中. 若是新的interval start 大于 heap peek, 就一直heap poll.
maxOverlap是这个过程中heap size 的峰值。
Time Complexity: O(nlogn). Space: O(n).
AC Java:
1 /** 2 * Definition for an interval. 3 * public class Interval { 4 * int start; 5 * int end; 6 * Interval() { start = 0; end = 0; } 7 * Interval(int s, int e) { start = s; end = e; } 8 * } 9 */ 10 public class Solution { 11 public int minMeetingRooms(Interval[] intervals) { 12 if(intervals == null || intervals.length == 0){ 13 return 0; 14 } 15 Arrays.sort(intervals, new Comparator<Interval>(){ 16 public int compare(Interval i1, Interval i2){ 17 if(i1.start == i2.start){ 18 return i1.end - i2.end; 19 } 20 return i1.start - i2.start; 21 } 22 }); 23 24 int maxOverlap = 0; 25 //Min Heap 26 PriorityQueue<Integer> que = new PriorityQueue<Integer>(); 27 for(int i = 0; i<intervals.length; i++){ 28 que.add(intervals[i].end); 29 while(!que.isEmpty() && intervals[i].start >= que.peek()){ 30 que.poll(); 31 } 32 maxOverlap = Math.max(maxOverlap, que.size()); 33 } 34 return maxOverlap; 35 } 36 }
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原文地址：http://www.cnblogs.com/Dylan-Java-NYC/p/5206676.html
