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
class Solution { public int busyStudent(int[] startTime, int[] endTime, int queryTime) { Map<Integer, Integer> map = new HashMap(); int le = startTime.length; int res = 0; for(int i = 0; i < le; i++){ if(startTime[i] <= queryTime) map.put(i, startTime[i]); } List<Integer> list = new ArrayList(map.keySet()); for(int i = 0; i < list.size(); i++){ if(endTime[(int)list.get(i)] >= queryTime) res++; } return res; } }
要求:startTime必须小于等于queryTime,endTime必须大于等于queryTime即可