Java 算法-查找循环世界中重叠间隔的持续时间(24小时)
我一直在尝试找出两个时间范围之间重叠小时数的算法,例如: 应该12点回来 及 应该返回4 因此,请帮助我填补创建以下函数的空白:Java 算法-查找循环世界中重叠间隔的持续时间(24小时),java,algorithm,intervals,overlap,Java,Algorithm,Intervals,Overlap,我一直在尝试找出两个时间范围之间重叠小时数的算法,例如: 应该12点回来 及 应该返回4 因此,请帮助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
public static Long findOverlappingInterval(Long startTime1, Long endTime1,
Long startTime2, Long endTime2){
// Any suggestions?
}
谢谢
编辑:
我知道使用和
创建两个二进制数组并总结结果的解决方案。
意思是:
但这无助于我的具体需求,因为我想将算法的思想用于solr查询,因此使用数组和二进制运算符对我来说不是一个选项/**开始时间必须小于结束时间*/
/** 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;
}
公共静态int-findOverlappingInterval(int-startTime1、int-endTime1、int-startTime2、int-endTime2){
int重叠时间=0;
int[]time1=newint[Math.abs(endTime1-startTime1)];
对于(int i1=startTime1;i1
这个方法应该适用于你想要它做的事情。这对我有用。
不过我不确定包装部分
编辑
/**返回四个时间变量之间的重叠*/
公共静态int-findOverlappingInterval(int-startTime1、int-endTime1、int-startTime2、int-endTime2){
int重叠时间=0;
//第一次
int time1Length=0;
如果(结束时间1<开始时间1){
时间长度=24-起始时间1;
时间1长度+=结束时间1;
}
int[]时间1;
if(time1Length==0){
time1=newint[Math.abs(endTime1-startTime1)];
对于(int i1=startTime1;i1
这段新代码应该满足您的要求。它在24小时内循环。首先,对于间隔(a,b)
(a>b),我们可以很容易地将其分成两个间隔
(a , 23) and (0, b)
因此,问题变成了用
a查找(a,b)
和(a1,b1)
之间的重叠。您不需要创建数组,只需计算区间之间的交点即可。有三种情况:
没有分割的间隔
一个分开的间隔
两个间隔都是分开的
可以将拆分的间隔作为两个单独的间隔来解决。使用递归,您可以轻松做到这一点:
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);
}
}
}
public static Long find应用程序接口(Long startTime1、Long endTime1、Long startTime2、Long endTime2)
{
if(startTime1
检查此处的链接:
编辑:从以下链接插入代码:
import java.util.*;
导入java.lang.*;
导入java.io.*;
/*只有当类是公共的时,类的名称才必须是“Main”*/
表意文字
{
公共静态长分钟(长a、长b){
报税表((a)(b)?(a):(b));
}
公共静态void main(字符串[]args)引发java.lang.Exception
{
长s1=6L,e1=23L,s2=2L,e2=17L,ans=0L;
布尔brokern1,brokern2;
Brokern1=(s1在短时间内有许多令人惊讶的答案
我遵循了在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
,那么结果可以分解为两个单独的计算,范围是[s,24]
和[0,e]
<>这可以相互“做”,所以只有3个简单的情况需要考虑。
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);
}
}
}
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+"");
}
}
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));
}
}
......AAAAAAAAAAAAAAAAAA
..BBBBBBBBBBBBBBBB......
12
AAAAAAAAAAAAA...........
BBB.........BBBBBBBBBBBB
4
AAAAA......AAAAAAAAAAAAA
BBBB........BBBBBBBBBBBB
16
AAAAA.......AAAAAAAAAAAA
BBBB.......BBBBBBBBBBBBB
16