算法 - 查找循环世界中重叠间隔的持续时间（24 小时）

Question

我一直在尝试找出用于查找两个时间范围之间的重叠小时数的算法，例如：

应该return12.

和

应该return4.

所以请帮我填补创建以下函数的空白：

public static Long findOverlappingInterval(Long startTime1, Long endTime1,
                                           Long startTime2, Long endTime2){ 
    // Any suggestions?
}

谢谢。

编辑：我知道创建两个二进制数组的解决方案，使用 AND 并对结果求和。含义：

但这对我的特定需求没有帮助，因为我想将算法的思想用于 solr 查询，因此 使用数组和二元运算符对我来说不是一个选择 .

Answer 1

/** Start times must be smaller than end times */
public static int findOverlappingInterval(int startTime1, int endTime1, int startTime2, int endTime2) {
    int overlappingTime = 0;
    int[] time1 = new int[Math.abs(endTime1 - startTime1)];
    for (int i1 = startTime1; i1 < endTime1; i1++) {
        time1[i1 - startTime1] = i1;
    }
    int[] time2 = new int[Math.abs(endTime2 - startTime2)];
    for (int i2 = startTime2; i2 < endTime2; i2++) {
        time2[i2 - startTime2] = i2;
    }

    for (int i = 0; i < time1.length; i++) {
        for (int j = 0; j < time2.length; j++) {
            if (time1[i] == time2[j]) {
                overlappingTime++;
            }
        }
    }
    return overlappingTime;
}

此方法应该适用于您希望它执行的操作。它对我有用。不过我不确定包装部分。

编辑

/** Returns the overlap between the four time variables */
public static int findOverlappingInterval(int startTime1, int endTime1, int startTime2, int endTime2) {
    int overlappingTime = 0;

    // First time
    int time1Length = 0;
    if (endTime1 < startTime1) {
        time1Length = 24 - startTime1;
        time1Length += endTime1;
    }
    int[] time1;
    if (time1Length == 0) {
        time1 = new int[Math.abs(endTime1 - startTime1)];
        for (int i1 = startTime1; i1 < endTime1; i1++) {
            time1[i1 - startTime1] = i1;
        }
    } else {
        time1 = new int[time1Length];
        int count = 0;
        for (int i1 = 0; i1 < endTime1; i1++) {
            time1[count] = i1;
            count++;
        }
        for (int i1 = startTime1; i1 < 24; i1++) {
            time1[count] = i1;
            count++;
        }
    }

    // Second time
    int time2Length = 0;
    if (endTime2 < startTime2) {
        time2Length = 24 - startTime2;
        time2Length += endTime2;
    }
    int[] time2;
    if (time2Length == 0) {
        time2 = new int[Math.abs(endTime2 - startTime2)];
        for (int i2 = startTime2; i2 < endTime2; i2++) {
            time2[i2 - startTime2] = i2;
        }
    } else {
        time2 = new int[time2Length];
        int count = 0;
        for (int i2 = 0; i2 < endTime2; i2++) {
            time2[count] = i2;
            count++;
        }
        for (int i2 = startTime2; i2 < 24; i2++) {
            time2[count] = i2;
            count++;
        }
    }

    // Overlap calculator
    for (int i = 0; i < time1.length; i++) {
        for (int j = 0; j < time2.length; j++) {
            if (time1[i] == time2[j]) {
                overlappingTime++;
            }
        }
    }
    return overlappingTime;
}

这个新代码应该可以满足您的需求。它以 24 小时为周期循环。

Answer 2

首先，对于区间(a, b) which (a > b)，我们可以很容易的把它分成两个区间

(a , 23) and (0, b)

所以，问题变成了找到 (a , b) 和 (a1, b1) 与 a <= b and a1 <= b1

之间的重叠

所以，有四种情况：

b < a1 或 b1 < a ，表示没有重叠， return 0
a <= a1 && b1 <= b，即(a, b)包含(a1, b1)，return b1 - a1 + 1
a1 <= a && b <= b1，即(a1, b1)包含(a, b)，return b - a + 1
最后一种情况是(a, b)和(a1, b1)每个区间的重叠部分，即重叠区间是(max(a1, a), min(b, b1))

Answer 3

不用创建数组也可以做到，只计算区间之间的交集。一共有三种情况：

没有拆分间隔。
一个分割间隔。
两个间隔分开。

您可以将拆分间隔问题解决为两个单独的间隔。使用递归你可以很容易地做到这一点：

public static Long findOverlappingInterval(Long startTime1, Long endTime1, Long startTime2, Long endTime2)
{
    if (startTime1 < endTime1 && startTime2 < endTime2) 
        return Math.max(0, Math.min(endTime2, endTime1) - Math.max(startTime1, startTime2) + 1);
    else
    {
        if (startTime1 < endTime1) 
            return findOverlappingInterval(startTime1, endTime1, 0L, endTime2) + 
                   findOverlappingInterval(startTime1, endTime1, startTime2, 23L);
        else if (startTime2 < endTime2) 
            return findOverlappingInterval(0L, endTime1, startTime2, endTime2) + 
                   findOverlappingInterval(startTime1, 23L, startTime2, endTime2);
        else
        {
            return findOverlappingInterval(0L, endTime1, 0L, endTime2) +
                   findOverlappingInterval(0L, endTime1, startTime2, 23L) +
                   findOverlappingInterval(startTime1, 23L, 0L, endTime2) +
                   findOverlappingInterval(startTime1, 23L, startTime2, 23L);
        }
    }
}

Answer 4

在此处检查 link：Ideone link</code> <blockquote> EDIT: Inserted the code from the link here: </blockquote> <pre><code>import java.util.*; import java.lang.*; import java.io.*; /* Name of the class has to be "Main" only if the class is public. */ class Ideone { public static Long min(Long a,Long b){ return ((a)<(b)?(a):(b)); } public static Long max(Long a,Long b){ return ((a)>(b)?(a):(b)); } public static void main (String[] args) throws java.lang.Exception { Long s1=6L,e1=23L,s2=2L,e2=17L,ans=0L; Boolean broken1,broken2; broken1=(s1<=e1)?false:true; broken2=(s2<=e2)?false:true; if(broken1){ if(broken2) ans=min(e1,e2)+1 + 23-max(s1,s2)+1; else{ if(e1>=s2) ans+=(min(e1,e2)-s2+1); if(s1<=e2) ans+=(e2-max(s1,s2)+1); } } else{ if(broken2){ if(e2>=s1) ans+=(min(e1,e2)-s1+1); if(s2<=e1) ans+=(e1-max(s1,s2)+1); } else{ if(e1<s2 || e2<s1) ans=0L; else ans=min(e1,e2)-max(s1,s2)+1; } } System.out.println(ans+""); } }

Answer 5

_{令人惊讶的是，短时间内得到了很多答案...}

我遵循了其他答案中已经提出的相同想法：当开始时间 s 小于结束时间 e 时，结果可以分解为两个单独的计算，对于范围 [s,24] 和 [0,e].

这个可以做到"mutually"所以只考虑3个简单的情况，剩下的可以用递归调用来完成。

不过，我试过了

考虑到（根据图像）终点应该包含 (!)
再添加一些测试用例
很好地可视化配置:-)

这是 MCVE:

的结果

public class OverlappingIntervals
{
    private static final long INTERVAL_SIZE = 24;

    public static void main(String[] args)
    {
        test(6,23, 2,17);
        test(0,12, 12,2);

        test(11,4, 12,3);
        test(12,4, 11,3);
    }

    private static void test(
        long s0, long e0, long s1, long e1)
    {
        System.out.println(createString(s0, e0, s1, e1));
        System.out.println(findOverlappingInterval(s0, e0, s1, e1));
    }

    private static String createString(
        long s0, long e0, long s1, long e1)
    {
        StringBuilder sb = new StringBuilder();
        sb.append(createString(s0, e0, "A")).append("\n");
        sb.append(createString(s1, e1, "B"));
        return sb.toString();
    }

    private static String createString(long s, long e, String c)
    {
        StringBuilder sb = new StringBuilder();
        for (int i=0; i<INTERVAL_SIZE; i++)
        {
            if (s < e)
            {
                if (i >= s && i <= e)
                {
                    sb.append(c);
                }
                else
                {
                    sb.append(".");
                }
            }
            else 
            {
                if (i <= e || i >= s)
                {
                    sb.append(c);
                }
                else 
                {
                    sb.append(".");
                }
            }
        }
        return sb.toString();
    }



    public static long findOverlappingInterval(
        long s0, long e0, long s1, long e1)
    {
        return compute(s0, e0+1, s1, e1+1);
    }

    public static long compute(
        long s0, long e0, long s1, long e1)
    {
        if (s0 > e0)
        {
            return 
                compute(s0, INTERVAL_SIZE, s1, e1) +
                compute(0, e0, s1, e1);
        }
        if (s1 > e1)
        {
            return 
                compute(s0, e0, s1, INTERVAL_SIZE) +
                compute(s0, e0, 0, e1);
        }
        return Math.max(0, Math.min(e0, e1) - Math.max(s0, s1));
    }
}

前两个测试用例是问题中给出的测试用例，它们分别正确打印12和4。其余两个用于测试其他重叠配置：

......AAAAAAAAAAAAAAAAAA
..BBBBBBBBBBBBBBBB......
12
AAAAAAAAAAAAA...........
BBB.........BBBBBBBBBBBB
4
AAAAA......AAAAAAAAAAAAA
BBBB........BBBBBBBBBBBB
16
AAAAA.......AAAAAAAAAAAA
BBBB.......BBBBBBBBBBBBB
16

但是，请注意，可能必须创建进一步的测试配置才能涵盖所有可能的情况。

算法 - 查找循环世界中重叠间隔的持续时间（24 小时）

Algorithms - Find duration of overlapping intervals in a cyclic world (24 hours)

java

algorithm

overlap

intervals