我试图计算时间相交的最大值的计数。
我想要的预期结果来自下面的代码示例应为4.
总共有8次，共有6个值相交，一组4个，一组2个。
我想要得到的是交叉路口的最大值，但却无法让它发挥作用。
这是目前的代码。
void Main()
{
var times = new List<Times> {
new Times
{
Start = DateTime.Now,
End = DateTime.Now.AddMinutes(10)
},
new Times
{
Start = DateTime.Now,
End = DateTime.Now.AddMinutes(10)
},
new Times
{
Start = DateTime.Now.AddMinutes(2),
End = DateTime.Now.AddMinutes(10)
},
new Times
{
Start = DateTime.Now.AddMinutes(15),
End = DateTime.Now.AddMinutes(35)
},
new Times
{
Start = DateTime.Now.AddMinutes(25),
End = DateTime.Now.AddMinutes(42)
},
new Times
{
Start = DateTime.Now.AddMinutes(43),
End = DateTime.Now.AddMinutes(50)
},
new Times
{
Start = DateTime.Now.AddMinutes(55),
End = DateTime.Now.AddMinutes(89)
},
new Times
{
Start = DateTime.Now.AddMinutes(2),
End = DateTime.Now.AddMinutes(12)
}
};
times.OrderBy(x => x.Start);
var overlappingEvents =
(
from e1 in times
where times
.Where(e2 => e1 != e2)
.Where(e2 => e1.Start <= e2.End)
.Where(e2 => e1.End >= e2.Start)
.Any()
select e1).ToList();
overlappingEvents.OrderBy(x => x.Start);
overlappingEvents.Distinct().OrderBy(x => x.Start);
overlappingEvents.Distinct().OrderBy(x => x.Start).Count();
}
public class Times
{
public DateTime Start { get; set; }
public DateTime End { get; set; }
}
这是时间对象的输出
Start | End
05/04/2017 08:38:57 | 05/04/2017 08:48:57
05/04/2017 08:38:57 | 05/04/2017 08:48:57
05/04/2017 08:40:57 | 05/04/2017 08:48:57
05/04/2017 08:40:57 | 05/04/2017 08:50:57
05/04/2017 08:53:57 | 05/04/2017 09:13:57
05/04/2017 09:03:57 | 05/04/2017 09:20:57
05/04/2017 09:21:57 | 05/04/2017 09:28:57
05/04/2017 09:33:57 | 05/04/2017 10:07:57
This is the output of the overlapping object ( I have added the line where the intersect stops)
Start | End
05/04/2017 08:38:57 | 05/04/2017 08:48:57
05/04/2017 08:38:57 | 05/04/2017 08:48:57
05/04/2017 08:40:57 | 05/04/2017 08:48:57
05/04/2017 08:40:57 | 05/04/2017 08:50:57
---------------------------------------
05/04/2017 08:53:57 | 05/04/2017 09:13:57
05/04/2017 09:03:57 | 05/04/2017 09:20:57
如果有人能帮助我获得十字路口的最大值，我将非常感激。
由于
答案 0 :(得分：2)
试试这个。如果它们至少有一个共同点（即矿石间隔的开始等于另一个区间的结束），我假设两个区间重叠。
void Main()
{
var times = new List<Times> {
new Times
{
Start = DateTime.Now,
End = DateTime.Now.AddMinutes(10)
},
new Times
{
Start = DateTime.Now,
End = DateTime.Now.AddMinutes(10)
},
new Times
{
Start = DateTime.Now.AddMinutes(2),
End = DateTime.Now.AddMinutes(10)
},
new Times
{
Start = DateTime.Now.AddMinutes(15),
End = DateTime.Now.AddMinutes(35)
},
new Times
{
Start = DateTime.Now.AddMinutes(25),
End = DateTime.Now.AddMinutes(42)
},
new Times
{
Start = DateTime.Now.AddMinutes(43),
End = DateTime.Now.AddMinutes(50)
},
new Times
{
Start = DateTime.Now.AddMinutes(55),
End = DateTime.Now.AddMinutes(89)
},
new Times
{
Start = DateTime.Now.AddMinutes(2),
End = DateTime.Now.AddMinutes(12)
}
};
var overlaps = times.Select(t1 => times.Count(t2 => IsOverlapping(t1, t2))).Max();
}
bool IsOverlapping(Times t1, Times t2)
{
if (t1.Start >= t2.Start && t1.Start <= t2.End)
{
return true;
}
if (t1.End >= t2.Start && t1.End <= t2.End)
{
return true;
}
if (t2.Start >= t1.Start && t2.Start <= t1.End)
{
return true;
}
if (t2.End >= t1.Start && t2.End <= t1.End)
{
return true;
}
return false;
}
public class Times
{
public DateTime Start { get; set; }
public DateTime End { get; set; }
}
IsOverlapping类：，则可以简化 Times函数
bool IsOverlapping(Times t1, Times t2)
{
return t1.Contains(t2.Start) || t1.Contains(t2.End) || t2.Contains(t1.Start) || t2.Contains(t1.End);
}
public class Times
{
public DateTime Start { get; set; }
public DateTime End { get; set; }
public bool Contains(DateTime date)
{
return date >= Start && date <= End;
}
}
答案 1 :(得分：0)
再向类中添加一个属性
package javaapplication3;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.net.Authenticator;
import java.net.PasswordAuthentication;
import java.net.URL;
import java.net.URLConnection;
import java.nio.charset.Charset;
import org.json.JSONArray;
import org.json.JSONException;
public class JSONREST {
public static String callURL(String myURL) {
System.out.println("Requested URL:" + myURL);
StringBuilder sb = new StringBuilder();
URLConnection urlConn = null;
InputStreamReader in = null;
try {
URL url = new URL(myURL);
urlConn = url.openConnection();
if (urlConn != null) {
urlConn.setReadTimeout(60 * 1000);
}
if (urlConn != null && urlConn.getInputStream() != null) {
in = new InputStreamReader(urlConn.getInputStream(),
Charset.defaultCharset());
BufferedReader bufferedReader = new BufferedReader(in);
if (bufferedReader != null) {
int cp;
while ((cp = bufferedReader.read()) != -1) {
sb.append((char) cp);
}
bufferedReader.close();
}
}
in.close();
} catch (Exception e) {
throw new RuntimeException("Exception while calling URL:" + myURL, e);
}
return sb.toString();
}
public static void main(String[] args) {
String jsonString = callURL("MY URL");
System.out.println("\n\njsonString: " + jsonString);
try {
JSONArray jsonArray = new JSONArray(jsonString);
System.out.println("\n\njsonArray: " + jsonArray);
}
catch (JSONException e) {
e.printStackTrace();
}
}
}
计算差距
public class Times
{
public DateTime Start { get; set; }
public DateTime End { get; set; }
public TimeSpan Gap { get; set; }
}