如何遍历Hashtable，然后将对象添加到ArrayList？

Question

我正在制作家谱程序。 我一直试图找出如何将对象从哈希表复制到ArrayList（只是对象Person，而不是String）。 我有一个孩子的哈希表，它们是Person对象。 我想检查我在程序中正在使用的currentPerson是否有子项，如果有，将这些子项添加到ArrayList childList中。 我已经为此工作了几个小时，似乎无法弄清楚。

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可以解决NG中提到的问题。

但是，如果您还计划填充grandchildList，则需要更改存储数据的方式。 由于您是按名称存储东西，因此您永远无法真正弄清楚答案是否正确。 假设您的Person类具有按名称列出的子级列表，则以下代码将起作用。

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());
    }
  }
}

---

考虑这种情况：

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

从技术上讲，只有一个grandChild，但是您的算法可能会返回两个，这取决于您存储信息的方式。