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  • 1450. Number of Students Doing Homework at a Given Time

    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

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  • 原文地址:https://www.cnblogs.com/wentiliangkaihua/p/12908309.html
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