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Java 正在获取控制台上输入的值数_Java_Collections - Fatal编程技术网

Java 正在获取控制台上输入的值数

Java 正在获取控制台上输入的值数,java,collections,Java,Collections,在下面的代码中,我取了三组值(IScanner s=new Scanner(System.In)); System.out.println(“请输入3对出生-死亡对”); List dinoList=new ArrayList(); //int-dinoStrength=s.nextInt(); 而(s.hasnetint()) { int num1=s.nextInt(); int num2=s.nextInt(); if(num2==num1)//您的情况(这里我假设出生日期不应等于死亡日期

在下面的代码中,我取了三组值(
I
Scanner s=new Scanner(System.In));
System.out.println(“请输入3对出生-死亡对”);
List dinoList=new ArrayList();
//int-dinoStrength=s.nextInt();
而(s.hasnetint())
{
int num1=s.nextInt();
int num2=s.nextInt();
if(num2==num1)//您的情况(这里我假设出生日期不应等于死亡日期。)
打破
恐龙d=新恐龙(num1,num2);
添加(d);
}

您可以尝试这样的操作。

替换
for(int i=0;i@Peter)这没有帮助,我只是不断输入值,没有任何结果。在结束输入或数字流之前,您不会得到结果。您希望如何操作?(在这种情况下,您可以编写任何非数字的内容,如
end
public static void main(String[] args) {

  Scanner s = new Scanner(System.in);
  System.out.println("please enter 3 birth - death pairs");
  List<Dinosaur> dinoList = new ArrayList<Dinosaur>();
  // int dinoStrength=s.nextInt();


  for (int i = 0; i <3; i++) 
  {
    int num1 = s.nextInt();
    int num2 = s.nextInt();
    Dinosaur d = new Dinosaur(num1, num2);
    dinoList.add(d);

  }
  //System.out.println(dinoList);

  Collections.sort(dinoList);

  //System.out.println(dinoList);
  int maxCount = 0;
  List<String> ls=new ArrayList<String>();

  for (Dinosaur dino : dinoList) 
  {
    // System.out.println("start date" + dino.getStartDate());
    // System.out.println("end date"+ dino.getEndDate());
    int count = 0;
    for (Dinosaur dino2 : dinoList) {

      if (dino2.getStartDate() <= dino.getEndDate()
          && dino2.getEndDate() >= dino.getStartDate())
        count++;

    }
    //System.out.println(count);

    if (maxCount < count) {
      maxCount = count;
      ls.clear();
      ls.add(dino.getStartDate()+"-"+dino.getEndDate());
    }

    else if(maxCount==count)

    {
      ls.add(dino.getStartDate()+"-"+dino.getEndDate());
    }
  }
  //System.out.println(maxCount);

  //System.out.println(ls);

  System.out.println("Max no of Dinos alive at the same time :"+maxCount);
}
Scanner s = new Scanner(System.in);
System.out.println("please enter 3 birth - death pairs");
List<Dinosaur> dinoList = new ArrayList<Dinosaur>();
// int dinoStrength=s.nextInt();


while(s.hasNextInt())
{

int num1 = s.nextInt();
int num2 = s.nextInt();
if(num2 ==num1)//Your condition(Here I assume date of birth should not be equal to death date.)
break;
Dinosaur d = new Dinosaur(num1, num2);
dinoList.add(d);

}