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我需要遍历对象的 ArrayList 并查找对多个对象具有相同值的变量。如下例所示，我正在查看一个具有子类的 ArrayList。我想要做的就是找出房子或公寓是否与此示例代码共享相同的列表编号。我尝试使用双循环和带有方法（sameListingNum）的增强循环，但无法解决。

谢谢

// main client class
public static void main(String[] args) 
    {
      ArrayList<House> listings = new ArrayList();

      listings.add(new House(0001, 200000.00));
      listings.add(new House(0201, 200000.00));
      listings.add(new House(0001, 200000.00));
      listings.add(new House(0401, 200000.00));
      listings.add(new House(0031, 200000.00));
      listings.add(new Condo(0401, 200000.00, 4));
      listings.add(new Condo(0001, 120000.00, 3));
      listings.add(new Condo(0301, 220000.00, 2));
      listings.add(new Condo(0001, 130000.00, 3));
      listings.add(new Condo(0201, 130000.00, 3));

      for(House currentHouse: listings)
        System.out.println(currentHouse);
      for(int i=0; i<listings.size()-1; i++)
      {
        for(int j=i+1; j<listings.size(); j++)
        {

        }
      } 

// House Class
public class House 
{
  public int listingNum;
  public double price; 

  public House()
  {
    listingNum = 0; 
    price = 0.00; 
  }  
  public House(int newListingNum, double newPrice)
  {
    listingNum = newListingNum;
    price = newPrice; 
  }  
  public int getListingNum()
  {
    return listingNum;
  }  
  public double getPrice()
  {
    return listingNum; 
  }
  public String toString()
  {
    return ("Listing number: "+listingNum+", Price: "+price);
  }
  public boolean sameListingNum(Object other)
  {
    if(!(other instanceof House))
      return false; 
    else {
      House objHouse = (House)other;
      if(listingNum - objHouse.getListingNum() == 0)
      {
        System.out.println("Same listing numbers: "
        +listingNum+", "+objHouse.getListingNum());
        return true;
      }
      else 
        return false;
    }
  }
}

// Condo Class
public class Condo extends House
{
  public int connectedUnits;

  public Condo()
  {
    super();
    connectedUnits = 0; 
  }
  public Condo(int newListingNum, double newPrice, int newConUnits)
  {
    super(newListingNum, newPrice);
    connectedUnits = newConUnits;
  }

  public double getPrice()
  {
    return price;
  }
  public int getListingNum()
  {
    return listingNum;
  }
  public int getConnectedUnits()
  {
    return connectedUnits; 
  }
  public String toString()
  {
    return super.toString()+", Number of connected unints: "+connectedUnits;
  }

public boolean sameListingNum(Object other)
          {
            if(!(other instanceof House))
              return false; 
            else {
              House objHouse = (House)other;
              if(listingNum - objHouse.getListingNum() == 0)
              {
                System.out.println("Same listing numbers: "
                +listingNum+", "+objHouse.getListingNum());
                return true;
              }
              else 
                return false;
            }
        }
4

3 回答 3

1

您可以使用列表映射对其进行分组，例如 Map> 组。

然后，您遍历您的列表，并为每个房屋/公寓将其放入相同列表编号的组中。最后，您将获得一张地图，其中每个条目都有具有相同列表编号的所有房屋/公寓。

这是一个示例：

Map<Integer, List<House>> groups = new HashMap<Integer, List<House>>();
for (House house:listings) {
   List<House> group = groups.get(house.getListingNum());
   if (group == null) {
     group = new ArrayList<House>();
     groups.put(house.getListingNum(), group);
   }
   group.add(house);
}

for (Entry<Integer, List<House>> entry:groups.entrySet()) System.out.println("Listing Number "+entry.getKey()+" Houses/condos: "+entry.getValue());
于 2013-10-24T15:31:20.113 回答
1

其他两个答案都可以交替使用，您可以在 House 上实施可比较的...例如。

public class House implements Comparable<House> 

@Override
public int compareTo(final House o) {
    return listingNum - o.listingNum;
}

然后在你的主要方法中。对 Collection 进行排序并检查前一行是否始终具有相同的 Id。

    Collections.sort(listings);

    int previousListing = Integer.MIN_VALUE;

    for (House currentHouse : listings) {
        if (currentHouse.getListingNum() == previousListing){
            System.out.println("Duplicate for " + currentHouse.getListingNum());
        }
        previousListing = currentHouse.getListingNum();
    }

任你选。

于 2013-10-24T15:47:04.533 回答
0

试试这个：

foreach(House h in listings)
{
if(listings.Exists(p => p.sameListingNum(h)))
{
//do something
}
}
于 2013-10-24T15:36:28.417 回答