检查日期是否重叠并返回最大计数

Question

d1:      |----------|
d2:            |------|
d3:        |--------------|
d4:                         |----|
d5:   |----|

Answer 1

我的答案会考虑：

• 给定(d3,d5)不重叠=>重叠(d1,d3,d5)=2，因为在给定时间只有两个日期重叠。

import java.time.LocalDate;
import java.util.ArrayList;
import java.util.List;

class Event {
    LocalDate startDate; // inclusive
    LocalDate endDate; // inclusive

    Event(LocalDate st, LocalDate end) {
        this.startDate = st;
        this.endDate = end;
    }

    // Getters & Setters omitted
}

public class Main {
    public static void main(String[] args) {
        List<Event> events = new ArrayList<Event>();
        events.add(new Event(LocalDate.of(2019,1,1), LocalDate.of(2019,5,1))); // d1
        events.add(new Event(LocalDate.of(2019,3,1), LocalDate.of(2019,6,1))); // d2
        events.add(new Event(LocalDate.of(2019,2,1), LocalDate.of(2019,7,1))); // d3
        events.add(new Event(LocalDate.of(2019,8,1), LocalDate.of(2019,12,1))); // d4
        // d5 do not overlap d3
        events.add(new Event(LocalDate.of(2018,12,1), LocalDate.of(2019,1,31))); // d5

        Integer startDateOverlaps = events.stream().map(Event::getStartDate).mapToInt(date -> overlap(date, events)).max().orElse(0);
        Integer endDateOverlaps = events.stream().map(Event::getEndDate).mapToInt(date -> overlap(date, events)).max().orElse(0);

        System.out.println(Integer.max(startDateOverlaps, endDateOverlaps));
    }

    public static Integer overlap(LocalDate date, List<Event> events) {
        return events.stream().mapToInt(event -> (! (date.isBefore(event.startDate) || date.isAfter(event.endDate))) ? 1 : 0).sum();
    }
}

我们求和每个重叠的日期（甚至其本身，否则(d1，d2，d3)将只计算(d2，d3)用于d1检查）并测试每个startDate和enddate。