从给定的日期范围列表中查找所有重叠的日期范围 [英] To find all the overlapping date ranges from a given list of Date Ranges

问题描述

我有一个BookingDateRange的列表，其中BookingDateRange是：

I have a List of BookingDateRange where BookingDateRange is :

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

        //getters & setters of properties
   }

要求：

1. 我需要查找是否有任何日期重叠在BookingDate列表中说dateRangeList


2. 如果是，查找所有日期范围对重叠表示字符串列表示overlapDatePairs

示例1 ：

Input1：

dateRangeList [0] = 23 dec 2012- 27 dec 2012

dateRangeList[0] = 23 dec 2012- 27 dec 2012

dateRangeList [1] = 14 dec 2012 - 25 dec 2012

dateRangeList[1] = 14 dec 2012 - 25 dec 2012

dateRangeList [2] = 1 jan 2012 - 23 jan 2012

dateRangeList[2] = 1 jan 2012 - 23 jan 2012

Output1：

isOverlappingDates = true

isOverlappingDates = true

overlapDatePairs = [0_1]

overlappingDatePairs = [0_1]

示例2：

Input2：

dateRangeList [0] = 23 dec 2012- 27 dec 2012

dateRangeList[0] = 23 dec 2012- 27 dec 2012

dateRangeList [1] = 1 jan 2012 - 23 jan 2012

dateRangeList[1] = 1 jan 2012 - 23 jan 2012

Output2：

isOverlappingDates = false

isOverlappingDates = false

overlapDate Pairs = []

overlappingDatePairs = []

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） 。是否有更好的复杂性？无法达到更复杂的算法。

The complexity of this approach is O(n ^2). Is a better complexity possible ? Could not arrive at an algorithm with a better complexity.

在SO中遇到了一些解决方案。但是没有一个可以完全满足要求。

Did come across a few solutions at SO. But none of them could cater to the requirement completely.

谢谢，
Shikha

Thanks, Shikha

推荐答案

这是O（nlog（n）），或者显然如果有很多碰撞，它是O（碰撞次数）。一个我曾经工作的公司使用类似于这个面试问题的东西。

Here's O(nlog(n)), or obviously if there are lots of collisions, it's O(number of collisions). A company I used to work for used something similar to this as an interview question.

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;
}

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