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1604-警告一小时内使用相同员工卡大于等于三次的人(Alert Using Same Key-Card Three or More Times in a One Hour Period)
发表于:2021-12-03 | 分类: 中等
字数统计: 1k | 阅读时长: 5分钟 | 阅读量:

原文链接: https://leetcode-cn.com/problems/alert-using-same-key-card-three-or-more-times-in-a-one-hour-period

英文原文

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").

Example 3:

Input: keyName = ["john","john","john"], keyTime = ["23:58","23:59","00:01"]
Output: []

Example 4:

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

 

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.

中文题目

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。

通过代码

高赞题解

class Solution {
public:
	vector<string> alertNames(vector<string>& names, vector<string>& times) {
		unordered_map<string, vector<int>> mp;
		for (int i = 0; i < names.size(); ++i) {
			string hour = times[i].substr(0, 2);
			string min = times[i].substr(3, 2);
			int time = stoi(hour) * 60 + stoi(min);
			mp[names[i]].push_back(time);
		}

		vector<string> ans;
		for (auto&& [name, nums] : mp) { //左右值通用引用
			if (nums.size() <= 2) continue;
			sort(nums.begin(), nums.end());
			for (int i = 2; i < nums.size(); ++i) {
				if (nums[i] - nums[i - 2] <= 60) {
					ans.push_back(name);
					break;
				}
			}
		}
		sort(ans.begin(), ans.end());
		return ans;
	}
};

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