LeetCode contest 189 5412. 在既定时间做作业的学生人数
Table of Contents
一、中文版
给你两个整数数组 startTime（开始时间）和 endTime（结束时间），并指定一个整数 queryTime 作为查询时间。
已知，第 i 名学生在 startTime[i] 时开始写作业并于 endTime[i] 时完成作业。
请返回在查询时间 queryTime 时正在做作业的学生人数。形式上，返回能够使 queryTime 处于区间 [startTime[i], endTime[i]]（含）的学生人数。
示例 1：
输入：startTime = [1,2,3], endTime = [3,2,7], queryTime = 4 输出：1 解释：一共有 3 名学生。 第一名学生在时间 1 开始写作业，并于时间 3 完成作业，在时间 4 没有处于做作业的状态。 第二名学生在时间 2 开始写作业，并于时间 2 完成作业，在时间 4 没有处于做作业的状态。 第二名学生在时间 3 开始写作业，预计于时间 7 完成作业，这是是唯一一名在时间 4 时正在做作业的学生。
示例 2：
输入：startTime = [4], endTime = [4], queryTime = 4 输出：1 解释：在查询时间只有一名学生在做作业。
示例 3：
输入：startTime = [4], endTime = [4], queryTime = 5 输出：0
示例 4：
输入：startTime = [1,1,1,1], endTime = [1,3,2,4], queryTime = 7 输出：0
示例 5：
输入：startTime = [9,8,7,6,5,4,3,2,1], endTime = [10,10,10,10,10,10,10,10,10], queryTime = 5 输出：5
提示：
startTime.length == endTime.length
1 <= startTime.length <= 100
1 <= startTime[i] <= endTime[i] <= 1000
1 <= queryTime <= 1000
二、英文版
Given two integer arrays startTime and endTime and given an integer queryTime.
The ith student started doing their homework at the time startTime[i] and finished it at time endTime[i].
Return the number of students doing their homework at time queryTime. More formally, return the number of students where queryTime lays in the interval [startTime[i], endTime[i]] inclusive.
Example 1:
Input: startTime = [1,2,3], endTime = [3,2,7], queryTime = 4 Output: 1 Explanation: We have 3 students where: The first student started doing homework at time 1 and finished at time 3 and wasn't doing anything at time 4. The second student started doing homework at time 2 and finished at time 2 and also wasn't doing anything at time 4. The third student started doing homework at time 3 and finished at time 7 and was the only student doing homework at time 4.
Example 2:
Input: startTime = [4], endTime = [4], queryTime = 4 Output: 1 Explanation: The only student was doing their homework at the queryTime.
Example 3:
Input: startTime = [4], endTime = [4], queryTime = 5 Output: 0
Example 5:
Input: startTime = [9,8,7,6,5,4,3,2,1], endTime = [10,10,10,10,10,10,10,10,10], queryTime = 5 Output: 5
Constraints:
startTime.length == endTime.length
1 <= startTime.length <= 100
1 <= startTime[i] <= endTime[i] <= 1000
1 <= queryTime <= 1000
三、My answer
class Solution: def busyStudent(self, startTime: List[int], endTime: List[int], queryTime: int) -> int: res = 0 for i in range(len(startTime)): if queryTime >= startTime[i] and queryTime <= endTime[i]: res += 1 return res
四、解题报告
模拟题目。
遍历 startTime 和 endTime，判断 queryTime 是否在两个数值中间即可。
