如何遍歷Hashtable，然后將對象添加到ArrayList？

Question

我正在制作家譜程序。 我一直試圖找出如何將對象從哈希表復制到ArrayList（只是對象Person，而不是String）。 我有一個孩子的哈希表，它們是Person對象。 我想檢查我在程序中正在使用的currentPerson是否有子項，如果有，將這些子項添加到ArrayList childList中。 我已經為此工作了幾個小時，似乎無法弄清楚。

import java.util.ArrayList;
import java.util.HashSet;
import java.util.Hashtable;

public class FamilyInfo {
//////////// instance variables
// hash table that tells the father of a named person, if known
private Hashtable<String, Person> fathers;
// hash table that tells the mother of a named person, if known
private Hashtable<String, Person> mothers;
// hash table that tells the children of a named person, if any are known.
// In theory the father, mother and children tables should be kept consistent
private Hashtable<String, HashSet<Person>> children;


/**
 * constructor -- initializes instance variables
 */
public FamilyInfo() {
    // initialize everything to be empty collections of the appropriate type
    fathers = new Hashtable<String, Person>();
    mothers = new Hashtable<String, Person>();
    children = new Hashtable<String, HashSet<Person>>();
}

    public ArrayList<String> grandchildNames(Person currentPerson) {
    // return a dummied up name telling that the method is not implemented
    ArrayList<String> rtnVal = new ArrayList<String>();

    //Create an ArrayList that will hold the child 
    ArrayList<Person> childList = new ArrayList<Person>();

    //Create an ArrayList that will hold the granchildren
    ArrayList<Person> grandchildList = new ArrayList<Person>();

    //Add children to the child list
    if(children.get(currentPerson.getName()) != null)
    {
       //add the children to childList from the children hashtable
    }


    return rtnVal;
    }

Answer 1

使用ArrayList addAll方法。

childList.addAll(children.get(currentPerson.getName())

Answer 2

使用ArrayList addAll可以解決NG中提到的問題。

但是，如果您還計划填充grandchildList，則需要更改存儲數據的方式。 由於您是按名稱存儲東西，因此您永遠無法真正弄清楚答案是否正確。 假設您的Person類具有按名稱列出的子級列表，則以下代碼將起作用。

if (children.containsKey(currentPerson.getName()) {
  for (Person c : children.get(currentPerson.getName())) {
    childList.add(c);
    //Now assuming Person class keeps a list of children names
    if (c.getChildren() != null) {
        grandchildList.addAll(c.getChildren());
    }
  }
}

---

考慮這種情況：

[Root]-> [Parent-Bob]-> [Child-Sam]-> [GrandChild-Peter]
[Root]-> [Parent-Sam]-> [Child-Bob]

從技術上講，只有一個grandChild，但是您的算法可能會返回兩個，這取決於您存儲信息的方式。