原文链接: https://leetcode-cn.com/problems/number-of-students-doing-homework-at-a-given-time
英文原文
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 4:
Input: startTime = [1,1,1,1], endTime = [1,3,2,4], queryTime = 7 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
中文题目
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。 第三名学生在时间 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
通过代码
高赞题解
解题思路
这是我来力扣做的最简单的一道题
代码
int busyStudent(int* startTime, int startTimeSize, int* endTime, int endTimeSize, int queryTime)
{
int ans=0;
for(int i=0; i<startTimeSize; i++)
if(queryTime >= startTime[i] && queryTime <= endTime[i])
ans++;
return ans;
}
统计信息
通过次数 | 提交次数 | AC比率 |
---|---|---|
24655 | 30701 | 80.3% |
提交历史
提交时间 | 提交结果 | 执行时间 | 内存消耗 | 语言 |
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