【leetcode】1604. Alert Using Same KeyCard Three or More Times in a One Hour Period

题目如下：

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.

Example 1:

Input: keyName = ["daniel","daniel","daniel","luis","luis","luis","luis"], keyTime = ["10:00","10:40","11:00","09:00","11:00","13:00","15:00"]
Output: ["daniel"]
Explanation: "daniel" used the keycard 3 times in a one-hour period ("10:00","10:40", "11:00").

Example 2:

Input: keyName = ["alice","alice","alice","bob","bob","bob","bob"], keyTime = ["12:01","12:00","18:00","21:00","21:20","21:30","23:00"]
Output: ["bob"]
Explanation: "bob" used the keycard 3 times in a one-hour period ("21:00","21:20", "21:30").

Constraints:

• 1 <= keyName.length, keyTime.length <= 105
• keyName.length == keyTime.length
• keyTime[i] is in the format "HH:MM".
• [keyName[i], keyTime[i]] is unique.
• 1 <= keyName[i].length <= 10
• keyName[i] contains only lowercase English letters.

解题思路：使用字典，以名字作为key，时间用list的方式有序存储，依次把时间存入list中，如果一个时间的插入位置是i，判断 i-2~i，i-1~i+1，i ~ i+2三个区间的间隔是否小于一小时。

代码如下：

class Solution(object):
    def alertNames(self, keyName, keyTime):
        """
        :type keyName: List[str]
        :type keyTime: List[str]
        :rtype: List[str]
        """
        res = set()
        def convert(time):
            hour,min = time.split(':')
            return int(hour) * 60 + int(min)
        import bisect
        dic = {}
        for name,time in zip(keyName,keyTime):
            if name not in dic:
                dic[name] = [convert(time)]
            else:
                converted_time = convert(time)
                bisect.insort_left(dic[name],converted_time)
                inx = bisect.bisect_left(dic[name],converted_time)
                if inx - 2 >= 0 and dic[name][inx] - dic[name][inx-2] <= 60:
                    res.add(name)
                elif inx + 2 < len(dic[name]) and dic[name][inx+2] - dic[name][inx] <= 60:
                    res.add(name)
                elif inx -1 >= 0 and inx + 1 < len(dic[name]) and dic[name][inx+1] - dic[name][inx-1] <= 60:
                    res.add(name)
        return sorted(list(res))