Difficulty :
Medium
Java
演算法Algorithm
撰寫人KVJK_2125Fri, Mar 29, 2024 22:30
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.
翻譯:
給兩個整數陣列 startTime (開始時間)和 endTime (結束時間),並給一個整數 queryTime。
第 i 位學生在 startTime[i] 時開始寫作業,在 endTime[i] 時完成作業。
回傳在 queryTime 正在做作業的學生人數。更正式的說,回傳能夠使 queryTime 位於區間 [startTime[i], endTime[i]] 內的學生人數。
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.
startTime.length == endTime.length
1 <= startTime.length <= 100
1 <= startTime[i] <= endTime[i] <= 1000
1 <= queryTime <= 1000
Step1.
建立一個整數變數int ans = startTime.length
,設定成和開始時間一樣多。
Step2.
判斷從startTime
到endTime
中,必須經過queryTime
。
因此,在startTime > queryTime
或endTime < queryTime
時,均不經過queryTime
,故ans--
。維持直到計算結束,回傳ans
。