从给定的日期范围列表中查找所有重叠的日期范围

Question

我有一个BookingDateRange列表,其中BookingDateRange是:

    public class BookingDateRange {
        private Date fromDate;
        private Date toDate;

        //getters & setters of properties
   }

需求:

1. 我需要查找在DateDate列表中是否有任何日期重复说dateRangeList
2. 如果是,找到所有重叠的日期范围对,比如说List of String说overlapDatePairs

例1:

输入1:

dateRangeList [0] = 2012年12月23日至2012年12月27日

dateRangeList [1] = 2012年12月14日 - 2012年12月25日

dateRangeList [2] = 2012年1月1日 - 2012年1月23日

输出1:

isOverlappingDates = true

overlapDatePairs = [0_1]

例2:

输入2:

dateRangeList [0] = 2012年12月23日至2012年12月27日

dateRangeList [1] = 2012年1月1日 - 2012年1月23日

输出2:

isOverlappingDates = false

overlapDatePairs = []

我的解决方案

/**
 * Checks if any of the dates overlap.
 *
 * @param dateRangeList the date range list
 * @param overlappingDatePairs the overlapping date pairs where overlappingDatePair is stored in the format dateRange1_dateRange2
 * @return true, if any of the dates overlap.
 */

public static boolean isOverlappingDates(
            List<BookingDateRange> dateRangeList,
            List<String> overlappingDatePairs) {

    boolean isOverlap = false;

    for (int index1 = 0; index1 < dateRangeList.size(); index1++) {
        for (int index2 = index1 + 1; index2 < dateRangeList.size(); index2++) {

            // Overlap exists if (StartA <= EndB) and (EndA >= StartB)

            Date startA = dateRangeList.get(index1).getFromDate();
            Date endA = dateRangeList.get(index1).getToDate();
            Date startB = dateRangeList.get(index2).getFromDate();
            Date endB = dateRangeList.get(index2).getToDate();

            boolean isStartABeforeEndB = (startA.compareTo(endB)) < 0;
            boolean isEndAAfterStartB = (endA.compareTo(startB)) > 0;

            boolean isCurrentPairOverlap = false;

            isCurrentPairOverlap = isStartABeforeEndB && isEndAAfterStartB;

            if (isCurrentPairOverlap) {
                overlappingDatePairs.add(index1 + "_" + index2);
                isOverlap = true;
            }
        }

    }
    return isOverlap;

    }

这种方法的复杂性是O(n ^ 2).更复杂的可能吗？无法获得具有更好复杂性的算法.

在SO处遇到了一些解决方案.但它们都不能完全满足要求.

谢谢,Shikha

Answer 1

这是O(nlog(n)),或者显然如果有很多碰撞,那就是O(碰撞次数).我曾经工作过的一家公司使用类似于此的面试问题.

private static class BookingTuple implements Comparable<BookingTuple> {
    public final Date date;
    public final boolean isStart;
    public final int id;
    public BookingTuple(Date date, boolean isStart, int id) {
        this.date = date;
        this.isStart = isStart;
        this.id = id;
    }

    @Override
    public int compareTo(BookingTuple other) {
        int dateCompare = date.compareTo(other.date);
        if (dateCompare != 0) {
            return dateCompare;
        } else {
            if (!isStart && other.isStart) {
                return -1;
            } else if (isStart && !other.isStart) {
                return 1;
            } else {
                return 0;
            }
        }
    }
}

public static boolean isOverlappingDates(List<BookingDateRange> dateRangeList, List<String> overlappingDatePairs) {
    List<BookingTuple> list = new ArrayList<BookingTuple>();
    for (int i = 0; i < dateRangeList.size(); i++) {
        Date from = dateRangeList.get(i).getFromDate();
        Date to = dateRangeList.get(i).getToDate();
        list.add(new BookingTuple(from, true, i));
        list.add(new BookingTuple(to, false, i));
    }

    Collections.sort(list);

    boolean overlap = false;

    HashSet<Integer> active = new HashSet<Integer>();
    for (BookingTuple tuple : list) {
        if (!tuple.isStart) {
            active.remove(tuple.id);
        } else {
            for (Integer n : active) {
                overlappingDatePairs.add(n + "_" + tuple.id);
                overlap = true;
            }
            active.add(tuple.id);
        }
    }

    return overlap;
}