我有一个代表等级关系的平面数据,如下所示:
ID Name PID 0 A NULL 1 B 0 2 C 0 4 D 1 5 E 1 6 F 4 3 G 0
此表表示'数据表',其中PID表示父元素。 例如,在第一行中我们看到A有PID null而B有PID 0,这意味着B的父是A,因为0是A的ID,A是根元素,因为它没有PID 。类似地,C具有父A,因为C也具有PID 0,并且0是A的ID。
我创建了一个类RecordHolder
来表示上表。我还实现了方法processRecordHolder
public Map<String, List<String>> processRecordHolder()
返回的映射使用element作为键,并将后代节点的集合作为值保存。例如,映射中的第一项对应于元素A,它具有许多后代,而元素C没有后代。输出中成员的顺序并不重要。
public static void main(String[] args) {
RecordHolder dt = new RecordHolder();
dt.addRow(0, "A", null);
dt.addRow(1, "B", 0);
dt.addRow(2, "C", 0);
dt.addRow(4, "D", 1);
dt.addRow(5, "E", 1);
dt.addRow(6, "F", 4);
dt.addRow(3, "G", 0);
System.out.println("Output:");
System.out.println(dt.processRecordHolder());
}
Output:
{D=[F], A=[B, C, G, D, E, F], B=[D, E, F]}
or
{D=[F], E=null, F=null, G=null, A=[B, C, G, D, E, F], B=[D, E, F], C=null}
以下是我迄今为止能够提出的Record
的实现:
public class Record {
public Integer id;
public String name;
public Integer parentId;
public Record parent;
public Collection<Record> children;
public Record(Integer id, String name, Integer parentId) {
this();
this.id = id;
this.name = name;
this.parentId = parentId;
}
public Record() {
children = Collections.newSetFromMap(new ConcurrentHashMap<Record, Boolean>())
}
public Collection<Record> getChildren() {
return children;
}
public Record getParent() {
return parent;
}
public Integer getParentId() {
return parentId;
}
@Override
public String toString() {
return "Record{" + "id=" + id + ", name=" + name + ", parentId=" + parentId + '}';
}
/* (non-Javadoc)
* @see java.lang.Object#hashCode()
*/
@Override
public int hashCode() {
final int prime = 31;
int result = 1;
result = prime * result + ((id == null) ? 0 : id.hashCode());
result = prime * result + ((name == null) ? 0 : name.hashCode());
result = prime * result + ((parentId == null) ? 0 : parentId.hashCode());
return result;
}
/* (non-Javadoc)
* @see java.lang.Object#equals(java.lang.Object)
*/
@Override
public boolean equals(Object obj) {
if (this == obj) {
return true;
}
if (obj == null) {
return false;
}
if (!(obj instanceof Record)) {
return false;
}
Record other = (Record) obj;
if (id == null) {
if (other.id != null) {
return false;
}
} else if (!id.equals(other.id)) {
return false;
}
if (name == null) {
if (other.name != null) {
return false;
}
} else if (!name.equals(other.name)) {
return false;
}
if (parentId == null) {
if (other.parentId != null) {
return false;
}
} else if (!parentId.equals(other.parentId)) {
return false;
}
return true;
}
}
现在我无法理解其他步骤该怎么办?
答案 0 :(得分:1)
如果您想尝试我更简单的实施想法,请在此处详细介绍。这样,您可以决定是否要使用当前的想法或尝试重新使用此想法。 (注意下面的代码是伪Java大纲,它不会编译,也不会测试):
int numNodes = 7;
Node[] nodes = new Node[numNodes];
//Read in your file here using a Scanner/FileReader or something
int ID = 0;
char value = 0;
int PID = 0;
while(scanner.hasNextLine()){
ID = scan.next();
value = scan.next();
PID = scan.next();
nodes[ID] = new Node(value, PID);
}
然后是节点类:
class Node{
char value;
Node parent;
public Node(value, parentID){
this.value = value;
if(parentID == -1)
parent = null;
else
parent = nodes[parentID]; //nodes will have to be a global array or get passed to the constructor
}
}
请注意,只有先前已初始化节点[parentID]中的项目时,此构造函数才有效。 (这是您当前输入文件顺序的情况,但可能不适用于其他情况。)
要使用此方法使用ID查找节点的祖先,只需执行以下操作:
printAncestry(nodes[ID]);
void printAncestry(Node n){
System.out.println("Child: " + n.value);
System.out.println("Ancestry: ");
while(n.parent != null){
n = n.parent;
System.out.println(n.value);
}
}
答案 1 :(得分:1)
尝试:
public class RecordHolder {
Map<Integer,String> namesById = new HashMap<>();
Map<Integer,List<Integer>> childrenById = new HashMap<>();
public void addRow(Integer id, String name, Integer parent) {
namesById.put(id, name);
List<Integer> children = childrenById.get(parent);
if (children == null) {
children = new ArrayList<>();
childrenById.put(parent, children);
}
children.add(id);
}
public Map<String,List<String>> processRecordHolder() {
Map<String,List<String>> results = new HashMap<>();
descendants(null, results);
return results;
}
private List<String> descendants(Integer id, Map<String, List<String>> results) {
final List<String> childrenNames = new ArrayList<>();
final List<Integer> childrenIds = childrenById.get(id);
if (childrenIds != null && childrenIds.size() > 0) {
for (Integer childrenId : childrenIds) {
final String childName = namesById.get(childrenId);
childrenNames.add(childName);
final List<String> grandchildrenNames = descendants(childrenId, results);
childrenNames.addAll(grandchildrenNames);
}
if (id != null) {
results.put(namesById.get(id), childrenNames);
}
}
return childrenNames;
}
}