本文介绍了java - 如何从linkedlist中删除节点?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
此代码是一个包含Inert name,delete,show和quit选项的表。
This code is a table that has an option to Inert name, delete, show, and quit .
此代码运行良好但我唯一的问题是如何删除节点中的选定名称
this code run's well but my only problem is on how to delete a chosen name in a node
class Node{
Node in;
String name;
public Node(){
in = null;
}
public Node(String n){
in = null;
name = n;
}
public void setIn(Node n){
in = n;
}
public Node getIn(){
return in;
}
public void setName(String n){
name = n;
}
public String getName(){
return name;
}
public class Main{
public static void main(String args[]){
Scanner scan = new Scanner(System.in);
LinkedList bi = new LinkedList();
while(true){
System.out.println("Choose!\n[a] Insert Name\n[b] Delete\n[c] Show\n[d] Exit");
char c = scan.next().charAt(0);
System.out.println();
if(c == 'a'){
System.out.print("Enter Name: ");
bi.insert(scan.next());
System.out.println();
}
else if(c == 'b'){
System.out.print("Enter Name to delete: ");
bi.delete(scan.next());
System.out.println();
}
else if(c == 'c'){
bi.show();
System.out.println();
}
else if(c == 'd'){
System.exit(0);
}
}
}
}
class LinkedList{
private Node root;
public LinkedList(){
root = null;
}
public void insert(String n){
root = insert(root, n);
}
private Node insert(Node n, String r){
if(n == null){
n = new Node(r);
}
else{
n.in = insert(n.in, r);
}
return n;
}
public void delete(String n){
root = delete(root, n);
}
private Node delete(Node n, String r){
}
public void show(){
show(root);
}
private Node show(Node n){
if(n == null){
System.out.println("Empy list!");
}
else{
while(n!=null){
System.out.println(n.getName());
n = n.getIn();
}
}
return n;
}
}
*我不知道如何删除节点。我应该把什么放在我的删除方法上?
*i don't know how to delete a node . What should i put on my delete method?
public void delete(String n){
root = delete(root, n);
}
private Node delete(Node n, String r){
}
推荐答案
要删除Node,您实际需要更新它的上一个节点要删除Node,最左边的Node最终会被垃圾收集。
To delete Node you actually need to update it's previous node's in to be deleting Node's in, and the left alone Node will eventually get garbage collected.
如果要删除的节点是根节点而不是更新根节点,则只有一个捕获。 / p>
Just one catch if node to be deleted is the root node than update root node.
private Node delete(Node root, String data)
{
Node n=null;
//in case list is empty then return
if(root==null) return n;
if (root.name.equals(data))
{
n = root;
root = null;
return n;
}
while(root.in!=null)
{
if (root.in.name.equals(data))
{
//save the reference
n=root.in;
//making root.in to be garbage collected
root.in = root.in.in;
break;
}
root = root.in;
}
return n;
}
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