【每日一题Day111】LC1604警告一小时内使用相同员工卡大于等于三次的人

给定员工姓名和使用工卡的时间戳列表，该程序旨在找出在一小时内使用工卡三次或更多次的员工，并按字母顺序返回他们的名字。通过使用哈希表存储每个员工的时间戳并进行排序，然后用双指针滑动窗口方法检查60分钟内的使用次数来实现这一目标。

警告一小时内使用相同员工卡大于等于三次的人【LC1604】

LeetCode company workers use key-cards to unlock office doors. Each time a worker uses their key-card, the security system saves the worker’s name and the time when it was used. The system emits an alert if any worker uses the key-card three or more times in a one-hour period.

You are given a list of strings keyName and keyTime where [keyName[i], keyTime[i]] corresponds to a person’s name and the time when their key-card was used in a single day.

Access times are given in the 24-hour time format “HH:MM”, such as "23:51" and "09:49".

Return a list of unique worker names who received an alert for frequent keycard use. Sort the names in ascending order alphabetically.

Notice that "10:00" - "11:00" is considered to be within a one-hour period, while "22:51" - "23:52" is not considered to be within a one-hour period.

2023/2/7

思路：
- 使用哈希表统计每位员工使用工卡的所有时间节点，将时间转化为分钟的形式，方便后续统计一小时内使用次数大于等于3的员工
- 然后将每位员工的时间节点从小到大排序，如果在60分钟的间隔内，有大于等于3个时间节点，那么需要对该员工发出警告
  - 双指针滑动窗口统计
  - 枚举每一个时间节点 $i$ ，判断其与第 $i - 1$ 个时间节点的差值是否小于等于60
- 返回结果需要按照字典顺序排序

实现

class Solution {
    public List<String> alertNames(String[] keyName, String[] keyTime) {
        Map<String,List<Integer>> map = new HashMap<>();;
        List<String> res = new LinkedList<>();
        int n = keyName.length;        
        for (int i = 0; i < n; i++){
            String name = keyName[i];
            String[] times = keyTime[i].split(":");
            int hour = Integer.parseInt(times[0]);
            int m = Integer.parseInt(times[1]);
            int t = hour * 60 + m;
            List<Integer> list = map.getOrDefault(name, new ArrayList<>());
            list.add(t);
            map.put(name, list);
        }
        for (var node : map.entrySet()){
            String name = node.getKey();
            List<Integer> times = node.getValue();
            int size = times.size();            
            Collections.sort(times);
            // 滑动窗口
            // if (size < 3) continue;
            // int l = 0, r = 0;
            // while (l < size && r < size){
            //     while (r < size && times.get(r) - times.get(l) <= 60 ){
            //         r++;
            //     } 
            //     if (r - l >= 3){
            //         res.add(name);
            //         break;
            //     }
            //     while (l < size && r < size && times.get(r) - times.get(l) > 60){
            //         l++;
            //     } 
            // }
            // 枚举
            for (int i = 2; i < size; i++){
                if (times.get(i) - times.get(i - 2) <= 60){
                    res.add(name);
                    break;
                }
            }
        }
        Collections.sort(res);
        return res;
    }
}