1353. 最多可以参加的会议数目

Posted 2021-05-15 lgz0921

链接：https://leetcode-cn.com/problems/maximum-number-of-events-that-can-be-attended/

思路：按照开始时间排序，然后从第一天开始循环遍历，把开始时间小于等于nowDay的结束时间放入优先队列中（小于等于nowDay说明有机会参加但是不一定能参加），优先队列是一个小顶堆，最上面是结束时间最早的那个，然后poll掉这个result++，最后天数后移一天nowDay++

上代码：

java实现：

class Solution {
    public int maxEvents(int[][] events) {
        Arrays.sort(events, Comparator.comparing(x -> x[0]));
        PriorityQueue<Integer> pQ = new PriorityQueue<>();
        int result = 0, nowDay = 1, i = 0;
        while (i < events.length || !pQ.isEmpty()) {
            while (!pQ.isEmpty() && pQ.peek() < nowDay) {
                pQ.poll();
            }
            while (i < events.length && events[i][0] <= nowDay){
                pQ.add(events[i++][1]);
            }
            if(!pQ.isEmpty()){
                pQ.poll();
                result++;
            }
            nowDay++;
        }
        return result;
    }
}

kotlin实现：

class Solution {
    fun maxEvents(events: Array<IntArray>): Int {
        events.sortBy { it[0] }
        val pQ = PriorityQueue<Int>()
        var result = 0
        var nowDay = 1
        var i = 0
        while (i < events.size || !pQ.isEmpty()){
            while (!pQ.isEmpty() && pQ.peek() < nowDay){
                pQ.poll()
            }
            while (i < events.size && events[i][0] <= nowDay ){
                pQ.add(events[i++][1])
            }
            if(!pQ.isEmpty()){
                pQ.poll()
                result++
            }
            nowDay++
        }
        return result
    }
}