2021-01-28 253. Meeting Rooms II

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该博客介绍了一个中等难度的算法问题，涉及堆、贪婪算法和排序。给定一系列会议时间区间，目标是确定所需的最少会议室数量。通过排序会议开始时间并使用优先队列来管理结束时间，可以有效地解决这个问题。示例展示了如何用Java实现这个解决方案。

253. Meeting Rooms II

Difficulty: Medium

Related Topics: Heap, Greedy, Sort

Given an array of meeting time intervals intervals where intervals[i] = [starti, endi], return the minimum number of conference rooms required.

Example 1:

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

Example 2:

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

Constraints:

1 <= intervals.length <= 104
0 <= starti < endi <= 106

Solution

Language: Java

class Solution {
    public int minMeetingRooms(int[][] intervals) {
        Comparator<int[]> comp = new Comparator<>() {
            public int compare(int[] a1, int[] a2) {
                return a1[0] - a2[0];
            }
        };
        
        if (intervals.length == 0) {
            return 0;
        }
        
        Arrays.sort(intervals, comp);
        
        PriorityQueue<Integer> pq = new PriorityQueue<>(new Comparator<Integer>() {
              public int compare(Integer a, Integer b) {
                return a - b;
              }
            });
        
        pq.add(intervals[0][1]);
        
        for (int i = 1; i < intervals.length; i++) {
            if (intervals[i][0] >=pq.peek()) {
                pq.poll();
            }
            pq.add(intervals[i][1]);
        }
        
        return pq.size();
    }
}