此代码是一个可以选择惰性名称,删除,显示和退出的表格。java - 如何从链表中删除节点?
这个代码运行的很好,但我唯一的问题是如何在一个节点
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;
}
}
*我不知道如何删除一个节点删除一个选择的名称。我应该把我的删除方法?
public void delete(String n){
root = delete(root, n);
}
private Node delete(Node n, String r){
}
正如其他人所说,先试一试。提示:在搜索要删除的节点时,使用两个指针遍历列表。 – Inertiatic