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;
}
答案 0 :(得分:0)
使用ArrayList addAll方法。
childList.addAll(children.get(currentPerson.getName())
答案 1 :(得分:0)
使用ArrayList addAll可以解决NG中提到的单行问题。
但是,如果您计划填写孙子列表,那么您需要更改存储数据的方式。由于您按名称存储内容,因此您无法确定答案是否正确。下面的代码将起作用,假设您的Person类具有名称的子列表。
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] - &gt; [家长 - 鲍勃] - &gt; [孩子 - 山姆] - &gt; [GrandChild - 彼得]
[根] - &gt; [家长 - 山姆] - &gt; [孩子 - 鲍勃]
技术上只有一个grandChild,但根据您存储信息的方式,您的算法可能会返回两个。