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📜 数据结构示例-删除循环链表中间节点
📅 最后修改于: 2020-10-15 06:02:44 🧑 作者: Mango
问：程序从循环链表的中间删除一个新节点。
说明
在此程序中，我们将创建一个循环链接列表，并从列表中间删除一个节点。如果列表为空，则显示消息“列表为空”。如果列表不为空，我们将计算列表的大小，然后将其除以2得到列表的中点。我们保持两个指针temp和current。当前将指向温度的前一个节点。我们将遍历列表直到到达中间点，然后当前点将指向中间节点。我们删除中间节点，使当前的下一个节点成为临时的下一个节点。
从列表中间删除节点后的循环链接列表
考虑上面的列表。列表的大小为4。节点的中点为2。要从列表中删除C，我们将遍历列表直到中点。现在，电流将指向B，温度将指向C。当B指向D时，C将被删除。
算法
- 定义一个Node类，该类代表列表中的一个节点。它具有两个属性数据，下一个将指向下一个节点。
- 定义另一个用于创建循环链接列表的类，它具有两个节点：head和tail。它具有可变大小和两种方法：deleteMid()和display()。
- deleteMid()将从列表中间删除该节点：
- 它首先检查head为空(空列表)，然后，由于列表中没有节点，它将从函数返回。
- 如果列表不为空，它将检查列表是否只有一个节点。
- 如果列表只有一个节点，则它将head和tail都设置为null。
- 如果列表具有多个节点，则它将计算列表的大小。将大小除以2并将其存储在变量计数中。
- Temp将指向头，当前将指向前一个节点。
- 迭代列表，直到当前将指向列表的中间节点为止。
- 当前将指向temp旁边的节点，即删除当前附近的节点。
- display()将显示列表中存在的所有节点。
- 定义一个新节点“当前”，该节点将指向头部。
- 打印current.data，直到current再次指向head。
- 当前将在每次迭代中指向列表中的下一个节点。
示例：
Python
#Represents the node of list.
class Node:
def __init__(self,data):
self.data = data;
self.next = None;
class CreateList:
#Declaring head and tail pointer as null.
def __init__(self):
self.head = Node(None);
self.tail = Node(None);
self.head.next = self.tail;
self.tail.next = self.head;
self.size = 0;
#This function will add the new node at the end of the list.
def add(self,data):
newNode = Node(data);
#Checks if the list is empty.
if self.head.data is None:
#If list is empty, both head and tail would point to new node.
self.head = newNode;
self.tail = newNode;
newNode.next = self.head;
else:
#tail will point to new node.
self.tail.next = newNode;
#New node will become new tail.
self.tail = newNode;
#Since, it is circular linked list tail will point to head.
self.tail.next = self.head;
#Size will count the number of nodes in the list
self.size = int(self.size)+1;
#Deletes node from the middle of the list
def deleteMid(self):
#Checks whether list is empty
if(self.head == None):
return;
else:
#Store the mid position of the list
count = (self.size//2) if (self.size % 2 == 0) else ((self.size+1)//2);
#Checks whether head is equal to tail or not, if yes then list has only one node.
if( self.head != self.tail ):
#Initially temp will point to head;
temp = self.head;
current = None;
#Current will point to node previous to temp
#If temp is pointing to node 2 then current will points to node 1.
for i in range(0, count-1):
current = temp;
temp = temp.next;
if(current != None):
#temp is the middle that needs to be removed.
#So, current node will point to node next to temp by skipping temp.
current.next = temp.next;
#Delete temp;
temp = None;
#Current points to null then head and tail will point to node next to temp.
else:
self.head = self.tail = temp.next;
self.tail.next = self.head;
#Delete temp;
temp = None;
#If the list contains only one element
#then it will remove it and both head and tail will point to null
else:
self.head = self.tail = None;
self.size = self.size - 1;
#Displays all the nodes in the list
def display(self):
current = self.head;
if self.head is None:
print("List is empty");
return;
else:
#Prints each node by incrementing pointer.
print(current.data),
while(current.next != self.head):
current = current.next;
print(current.data),
print("\n");
class CircularLinkedList:
cl = CreateList();
#Adds data to the list
cl.add(1);
cl.add(2);
cl.add(3);
cl.add(4);
#Printing original list
print("Original List:");
cl.display();
while(cl.head != None):
cl.deleteMid();
#Printing updated list
print("Updated List:");
cl.display();
输出：
Original List:
1 2 3 4
Updated List:
1 3 4
Updated List:
1 4
Updated List:
4
Updated List:
List is empty
C
#include
#include
#include
//Represents the node of list.
struct node{
int data;
struct node *next;
};
int size;
//Declaring head and tail pointer as null.
struct node *head = NULL;
struct node *tail = NULL;
//This function will add the new node at the end of the list.
void add(int data){
//Create new node
struct node *newNode = (struct node*)malloc(sizeof(struct node));
newNode->data = data;
//Checks if the list is empty.
if(head == NULL){
//If list is empty, both head and tail would point to new node.
head = newNode;
tail = newNode;
newNode->next = head;
}else {
//tail will point to new node.
tail->next = newNode;
//New node will become new tail.
tail = newNode;
//Since, it is circular linked list tail will point to head.
tail->next = head;
}
//Counts the number of nodes in list
size++;
}
//Deletes node from the middle of the list
void deleteMid() {
struct node *current, *temp;
//Checks whether list is empty
if(head == NULL) {
return;
}
else {
//Store the mid position of the list
int count = (size % 2 == 0) ? (size/2) : ((size+1)/2);
//Checks whether head is equal to tail or not, if yes then list has only one node.
if( head != tail ) {
//Initially temp will point to head;
temp = head;
current = NULL;
//Current will point to node previous to temp
//If temp is pointing to node 2 then current will points to node 1.
for(int i = 0; i < count-1; i++){
current = temp;
temp = temp->next;
}
if(current != NULL) {
//temp is the middle that needs to be removed.
//So, current node will point to node next to temp by skipping temp.
current->next = temp->next;
//Delete temp;
temp = NULL;
}
//Current points to null then head and tail will point to node next to temp.
else {
head = tail = temp->next;
tail->next = head;
//Delete temp;
temp = NULL;
}
}
//If the list contains only one element
//then it will remove it and both head and tail will point to null
else {
head = tail = NULL;
}
}
size--;
}
//This function will display the nodes of circular linked list
void display(){
struct node *current = head;
if(head == NULL){
printf("List is empty");
}
else{
do{
//Prints each node by incrementing pointer.
printf("%d ", current->data);
current = current->next;
}while(current != head);
printf("\n");
}
}
int main()
{
//Adds data to the list
add(1);
add(2);
add(3);
add(4);
//Printing original list
printf("Original List:\n ");
display();
while(head != NULL) {
deleteMid();
//Printing updated list
printf("Updated List:\n ");
display();
}
return 0;
}
输出：
Original List:
1 2 3 4
Updated List:
1 3 4
Updated List:
1 4
Updated List:
4
Updated List:
List is empty
JAVA
public class DeleteMid {
//Represents the node of list.
public class Node{
int data;
Node next;
public Node(int data) {
this.data = data;
}
}
public int size;
//Declaring head and tail pointer as null.
public Node head = null;
public Node tail = null;
//This function will add the new node at the end of the list.
public void add(int data){
//Create new node
Node newNode = new Node(data);
//Checks if the list is empty.
if(head == null) {
//If list is empty, both head and tail would point to new node.
head = newNode;
tail = newNode;
newNode.next = head;
}
else {
//tail will point to new node.
tail.next = newNode;
//New node will become new tail.
tail = newNode;
//Since, it is circular linked list tail will point to head.
tail.next = head;
}
//Counts the number of nodes in list
size++;
}
//Deletes node from the middle of the list
public void deleteMid() {
Node current, temp;
//Checks whether list is empty
if(head == null) {
return;
}
else {
//Store the mid position of the list
int count = (size % 2 == 0) ? (size/2) : ((size+1)/2);
//Checks whether head is equal to tail or not, if yes then list has only one node.
if( head != tail ) {
//Initially temp will point to head;
temp = head;
current = null;
//Current will point to node previous to temp
//If temp is pointing to node 2 then current will points to node 1.
for(int i = 0; i < count-1; i++){
current = temp;
temp = temp.next;
}
if(current != null) {
//temp is the middle that needs to be removed.
//So, current node will point to node next to temp by skipping temp.
current.next = temp.next;
//Delete temp;
temp = null;
}
//Current points to null then head and tail will point to node next to temp.
else {
head = tail = temp.next;
tail.next = head;
//Delete temp;
temp = null;
}
}
//If the list contains only one element
//then it will remove it and both head and tail will point to null
else {
head = tail = null;
}
}
size--;
}
//Displays all the nodes in the list
public void display() {
Node current = head;
if(head == null) {
System.out.println("List is empty");
}
else {
do{
//Prints each node by incrementing pointer.
System.out.print(" "+ current.data);
current = current.next;
}while(current != head);
System.out.println();
}
}
public static void main(String[] args) {
DeleteMid cl = new DeleteMid();
//Adds data to the list
cl.add(1);
cl.add(2);
cl.add(3);
cl.add(4);
//Printing original list
System.out.println("Original List: ");
cl.display();
while(cl.head != null) {
cl.deleteMid();
//Printing updated list
System.out.println("Updated List: ");
cl.display();
}
}
}
输出：
Original List:
1 2 3 4
Updated List:
1 3 4
Updated List:
1 4
Updated List:
4
Updated List:
List is empty
C＃
using System;
namespace CircularLinkedList
{
public class Program
{
//Represents the node of list.
public class Node{
public T data;
public Node next;
public Node(T value) {
data = value;
next = null;
}
}
public class CreateList{
public int size = 0;
public Node head = null;
public Node tail = null;
//This function will add the new node at the end of the list.
public void add(T data){
//Create new node
Node newNode = new Node(data);
//Checks if the list is empty.
if(head == null){
head = newNode;
tail = newNode;
newNode.next = head;
}else{
//tail will point to new node.
tail.next = newNode;
//New node will become new tail.
tail = newNode;
//Since, it is circular linked list tail will point to head.
tail.next = head;
}
//Counts the number of nodes in the list
size++;
}
//Deletes node from the middle of the list
public void deleteMid() {
Node current, temp;
//Checks whether list is empty
if(head == null) {
return;
}
else {
//Store the mid position of the list
int count = (size % 2 == 0) ? (size/2) : ((size+1)/2);
//Checks whether head is equal to tail or not, if yes then list has only one node.
if( head != tail ) {
//Initially temp will point to head;
temp = head;
current = null;
//Current will point to node previous to temp
//If temp is pointing to node 2 then current will points to node 1.
for(int i = 0; i < count-1; i++){
current = temp;
temp = temp.next;
}
if(current != null) {
//temp is the middle that needs to be removed.
//So, current node will point to node next to temp by skipping temp.
current.next = temp.next;
//Delete temp;
temp = null;
}
//Current points to null then head and tail will point to node next to temp.
else {
head = tail = temp.next;
tail.next = head;
//Delete temp;
temp = null;
}
}
//If the list contains only one element
//then it will remove it and both head and tail will point to null
else {
head = tail = null;
}
}
size--;
}
//Displays all the nodes in the list
public void display(){
Node current = head;
if(head == null){
Console.WriteLine("List is empty");
}
else{
do{
//Prints each node by incrementing pointer.
Console.Write(" "+ current.data);
current = current.next;
}while(current != head);
Console.WriteLine();
}
}
}
public static void Main()
{
CreateList cl = new CreateList();
//Adds data to the list
cl.add(1);
cl.add(2);
cl.add(3);
cl.add(4);
//Printing original list
Console.WriteLine("Original List: ");
cl.display();
while(cl.head != null) {
cl.deleteMid();
//Printing updated list
Console.WriteLine("Updated List: ");
cl.display();
}
}
}
}
输出：
Original List:
1 2 3 4
Updated List:
1 3 4
Updated List:
1 4
Updated List:
4
Updated List:
List is empty
PHP：
data = $data;
$this->next = NULL;
}
}
class CreateList{
//Declaring head and tail pointer as null.
public $head;
public $tail;
function __construct(){
$this->head = NULL;
$this->tail = NULL;
$this->size = 0;
}
//This function will add the new node at the end of the list.
function add($data){
//Create new node
$newNode = new Node($data);
//Checks if the list is empty.
if($this->head == NULL){
//If list is empty, both head and tail would point to new node.
$this->head = $newNode;
$this->tail = $newNode;
$newNode->next = $this->head;
}
else{
//tail will point to new node.
$this->tail->next = $newNode;
//New node will become new tail.
$this->tail = $newNode;
//Since, it is circular linked list tail will point to head.
$this->tail->next = $this->head;
}
$this->size++;
}
//Deletes node from the middle of the list
function deleteMid() {
//Checks whether list is empty
if($this->head == NULL) {
return;
}
else {
//Store the mid position of the list
$count = ($this->size % 2 == 0) ? ($this->size/2) : (($this->size+1)/2);
//Checks whether head is equal to tail or not, if yes then list has only one node.
if( $this->head != $this->tail ) {
//Initially temp will point to head;
$temp = $this->head;
$current = NULL;
//Current will point to node previous to temp
//If temp is pointing to node 2 then current will points to node 1.
for($i = 0; $i < $count-1; $i++){
$current = $temp;
$temp = $temp->next;
}
if($current != NULL) {
//temp is the middle that needs to be removed.
//So, current node will point to node next to temp by skipping temp.
$current->next = $temp->next;
//Delete temp;
$temp = NULL;
}
//Current points to null then head and tail will point to node next to temp.
else {
$this->head = $this->tail = $temp->next;
$this->tail->next = $this->head;
//Delete temp;
$temp = NULL;
}
}
//If the list contains only one element
//then it will remove it and both head and tail will point to null
else {
$this->head = $this->tail = NULL;
}
}
$this->size--;
}
//Displays all the nodes in the list
function display() {
$current = $this->head;
if($this->head == NULL) {
echo "List is empty";
}
else {
do{
//Prints each node by incrementing pointer.
echo(" $current->data");
$current = $current->next;
}while($current != $this->head);
echo "
";
}
}
}
$cl = new CreateList();
//Adds data to the list
$cl->add(1);
$cl->add(2);
$cl->add(3);
$cl->add(4);
//Printing original list
echo "Original List:
";
$cl->display();
while($cl->head != NULL) {
$cl->deleteMid();
//Printing updated list
echo "Updated List:
";
$cl->display();
}
?>
输出：
Original List:
1 2 3 4
Updated List:
1 3 4
Updated List:
1 4
Updated List:
4
Updated List:
List is empty