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5516. 警告一小时内使用相同员工卡大于等于三次的人-map/优先队列-中等

作者:互联网

问题描述

力扣公司的员工都使用员工卡来开办公室的门。每当一个员工使用一次他的员工卡,安保系统会记录下员工的名字和使用时间。如果一个员工在一小时时间内使用员工卡的次数大于等于三次,这个系统会自动发布一个 警告 。

给你字符串数组 keyName 和 keyTime ,其中 [keyName[i], keyTime[i]] 对应一个人的名字和他在 某一天 内使用员工卡的时间。

使用时间的格式是 24小时制 ,形如 "HH:MM" ,比方说 "23:51" 和 "09:49" 。

请你返回去重后的收到系统警告的员工名字,将它们按 字典序升序 排序后返回。

请注意 "10:00" - "11:00" 视为一个小时时间范围内,而 "23:51" - "00:10" 不被视为一小时内,因为系统记录的是某一天内的使用情况。

 

示例 1:

输入:keyName = ["daniel","daniel","daniel","luis","luis","luis","luis"], keyTime = ["10:00","10:40","11:00","09:00","11:00","13:00","15:00"]
输出:["daniel"]
解释:"daniel" 在一小时内使用了 3 次员工卡("10:00","10:40","11:00")。
示例 2:

输入:keyName = ["alice","alice","alice","bob","bob","bob","bob"], keyTime = ["12:01","12:00","18:00","21:00","21:20","21:30","23:00"]
输出:["bob"]
解释:"bob" 在一小时内使用了 3 次员工卡("21:00","21:20","21:30")。
示例 3:

输入:keyName = ["john","john","john"], keyTime = ["23:58","23:59","00:01"]
输出:[]
示例 4:

输入:keyName = ["leslie","leslie","leslie","clare","clare","clare","clare"], keyTime = ["13:00","13:20","14:00","18:00","18:51","19:30","19:49"]
输出:["clare","leslie"]
 

提示:

1 <= keyName.length, keyTime.length <= 105
keyName.length == keyTime.length
keyTime 格式为 "HH:MM" 。
保证 [keyName[i], keyTime[i]] 形成的二元对 互不相同 。
1 <= keyName[i].length <= 10
keyName[i] 只包含小写英文字母。

来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period

解答

//朴实无华的map硬怼,用了下优先队列,这样就可以省去对每个人打卡时间的排序了
class Solution {
    public int minites(String time){
        return Integer.parseInt(time.substring(0,2))*60+Integer.parseInt(time.substring(3,5));
    }
    public List<String> alertNames(String[] keyName, String[] keyTime) {
        List<String> res = new ArrayList<>();
        if(keyName.length!=keyTime.length || keyName.length==0)return res;
        Map<String, Queue<Integer>> map = new HashMap<>();
        int count = 0;
        for(String name:keyName){
            if(!map.containsKey(name))map.put(name, new PriorityQueue<Integer>());
            map.get(name).offer(minites(keyTime[count]));
            count++;
        }
        for(String name:map.keySet()){
            Queue<Integer> hour = new LinkedList<>();
            Queue<Integer> temp = map.get(name);
            while(!temp.isEmpty()){
                int next = temp.poll();
                if(hour.isEmpty())hour.offer(next);
                else{
                    while(!hour.isEmpty() && next-hour.peek()>60){
                        hour.poll();
                    }
                    hour.offer(next);
                }
                if(hour.size()>=3){
                    res.add(name);
                    break;
                }
            }
        }
        Collections.sort(res);
        return res;
    }
}

 

标签:map,00,hour,队列,员工,keyName,keyTime,5516
来源: https://www.cnblogs.com/xxxxxiaochuan/p/13769663.html