Escriba una función en C que mueva el primer elemento hasta el final en una lista enlazada individual dada. Por ejemplo, si la lista enlazada dada es 1->2->3->4->5, entonces la función debería cambiar la lista a 2->3->4->5->1.
Algoritmo:
Recorra la lista hasta el último Node. Use dos punteros: uno para almacenar la dirección del último Node (último) y otro para la dirección del primer Node (primero). Después del final del ciclo, realice las siguientes operaciones.
i) Hacer head como segundo Node (*head_ref = first->next).
ii) Establecer el siguiente del primero como NULL (primero->siguiente = NULL).
iii) Establecer el penúltimo como primero (último->siguiente = primero)
C++
/* C++ Program to move first element to end
in a given linked list */
#include <stdio.h>
#include <stdlib.h>
/* A linked list node */
struct Node {
int data;
struct Node* next;
};
/* We are using a double pointer head_ref
here because we change head of the linked
list inside this function.*/
void moveToEnd(struct Node** head_ref)
{
/* If linked list is empty, or it contains
only one node, then nothing needs to be
done, simply return */
if (*head_ref == NULL || (*head_ref)->next == NULL)
return;
/* Initialize first and last pointers */
struct Node* first = *head_ref;
struct Node* last = *head_ref;
/*After this loop last contains address
of last node in Linked List */
while (last->next != NULL) {
last = last->next;
}
/* Change the head pointer to point
to second node now */
*head_ref = first->next;
/* Set the next of first as NULL */
first->next = NULL;
/* Set the next of last as first */
last->next = first;
}
/* UTILITY FUNCTIONS */
/* Function to add a node at the beginning
of Linked List */
void push(struct Node** head_ref, int new_data)
{
struct Node* new_node = new Node;
new_node->data = new_data;
new_node->next = (*head_ref);
(*head_ref) = new_node;
}
/* Function to print nodes in a given linked list */
void printList(struct Node* node)
{
while (node != NULL) {
printf("%d ", node->data);
node = node->next;
}
}
/* Driver program to test above function */
int main()
{
struct Node* start = NULL;
/* The constructed linked list is:
1->2->3->4->5 */
push(&start, 5);
push(&start, 4);
push(&start, 3);
push(&start, 2);
push(&start, 1);
printf("\n Linked list before moving first to end\n");
printList(start);
moveToEnd(&start);
printf("\n Linked list after moving first to end\n");
printList(start);
return 0;
}
Java
/* Java Program to move first element to end
in a given linked list */
class Sol
{
/* A linked list node */
static class Node
{
int data;
Node next;
};
/* We are using a double pointer head_ref
here because we change head of the linked
list inside this function.*/
static Node moveToEnd( Node head_ref)
{
/* If linked list is empty, or it contains
only one node, then nothing needs to be
done, simply return */
if (head_ref == null || (head_ref).next == null)
return null;
/* Initialize first and last pointers */
Node first = head_ref;
Node last = head_ref;
/*After this loop last contains address
of last node in Linked List */
while (last.next != null)
{
last = last.next;
}
/* Change the head pointer to point
to second node now */
head_ref = first.next;
/* Set the next of first as null */
first.next = null;
/* Set the next of last as first */
last.next = first;
return head_ref;
}
/* UTILITY FUNCTIONS */
/* Function to add a node at the beginning
of Linked List */
static Node push( Node head_ref, int new_data)
{
Node new_node = new Node();
new_node.data = new_data;
new_node.next = (head_ref);
(head_ref) = new_node;
return head_ref;
}
/* Function to print nodes in a given linked list */
static void printList( Node node)
{
while (node != null)
{
System.out.printf("%d ", node.data);
node = node.next;
}
}
/* Driver code */
public static void main(String args[])
{
Node start = null;
/* The constructed linked list is:
1.2.3.4.5 */
start = push(start, 5);
start = push(start, 4);
start = push(start, 3);
start = push(start, 2);
start = push(start, 1);
System.out.printf("\n Linked list before moving first to end\n");
printList(start);
start = moveToEnd(start);
System.out.printf("\n Linked list after moving first to end\n");
printList(start);
}
}
// This code is contributed by Arnab Kundu
Python3
# Python3 Program to move first element to end
# in a given linked list
# A linked list node
class Node:
def __init__(self):
self.data = 0
self.next = None
# We are using a double pointer head_ref
# here because we change head of the linked
# list inside this function.
def moveToEnd( head_ref) :
# If linked list is empty, or it contains
# only one node, then nothing needs to be
# done, simply return
if (head_ref == None or (head_ref).next == None) :
return None
# Initialize first and last pointers
first = head_ref
last = head_ref
# After this loop last contains address
# of last node in Linked List
while (last.next != None) :
last = last.next
# Change the head pointer to point
# to second node now
head_ref = first.next
# Set the next of first as None
first.next = None
# Set the next of last as first
last.next = first
return head_ref
# UTILITY FUNCTIONS
# Function to add a node at the beginning
# of Linked List
def push( head_ref, new_data) :
new_node = Node()
new_node.data = new_data
new_node.next = (head_ref)
(head_ref) = new_node
return head_ref
# Function to print nodes in a given linked list
def printList(node) :
while (node != None):
print(node.data, end = " ")
node = node.next
# Driver code
start = None
# The constructed linked list is:
#1.2.3.4.5
start = push(start, 5)
start = push(start, 4)
start = push(start, 3)
start = push(start, 2)
start = push(start, 1)
print("\n Linked list before moving first to end")
printList(start)
start = moveToEnd(start)
print("\n Linked list after moving first to end")
printList(start)
# This code is contributed by Arnab Kundu
C#
/* C# Program to move first element to end
in a given linked list */
using System;
class GFG
{
/* A linked list node */
public class Node
{
public int data;
public Node next;
};
/* We are using a double pointer head_ref
here because we change head of the linked
list inside this function.*/
static Node moveToEnd( Node head_ref)
{
/* If linked list is empty, or it contains
only one node, then nothing needs to be
done, simply return */
if (head_ref == null || (head_ref).next == null)
return null;
/* Initialize first and last pointers */
Node first = head_ref;
Node last = head_ref;
/*After this loop last contains address
of last node in Linked List */
while (last.next != null)
{
last = last.next;
}
/* Change the head pointer to point
to second node now */
head_ref = first.next;
/* Set the next of first as null */
first.next = null;
/* Set the next of last as first */
last.next = first;
return head_ref;
}
/* UTILITY FUNCTIONS */
/* Function to add a node at the beginning
of Linked List */
static Node push( Node head_ref, int new_data)
{
Node new_node = new Node();
new_node.data = new_data;
new_node.next = (head_ref);
(head_ref) = new_node;
return head_ref;
}
/* Function to print nodes in a given linked list */
static void printList( Node node)
{
while (node != null)
{
Console.Write("{0} ", node.data);
node = node.next;
}
}
/* Driver code */
public static void Main(String []args)
{
Node start = null;
/* The constructed linked list is:
1.2.3.4.5 */
start = push(start, 5);
start = push(start, 4);
start = push(start, 3);
start = push(start, 2);
start = push(start, 1);
Console.Write("\n Linked list before moving first to end\n");
printList(start);
start = moveToEnd(start);
Console.Write("\n Linked list after moving first to end\n");
printList(start);
}
}
// This code is contributed by 29AjayKumar
Javascript
<script>
// JavaScript program to move first element
// to end in a given linked list
/* A linked list node */
class Node
{
constructor()
{
this.data = 0;
this.next = null;
}
}
/* We are using a double pointer head_ref
here because we change head of the linked
list inside this function.*/
function moveToEnd(head_ref)
{
/* If linked list is empty, or it contains
only one node, then nothing needs to be
done, simply return */
if (head_ref == null || head_ref.next == null)
return null;
/* Initialize first and last pointers */
var first = head_ref;
var last = head_ref;
/*After this loop last contains address
of last node in Linked List */
while (last.next != null)
{
last = last.next;
}
/* Change the head pointer to point
to second node now */
head_ref = first.next;
/* Set the next of first as null */
first.next = null;
/* Set the next of last as first */
last.next = first;
return head_ref;
}
/* UTILITY FUNCTIONS */
/* Function to add a node at the beginning
of Linked List */
function push(head_ref, new_data)
{
var new_node = new Node();
new_node.data = new_data;
new_node.next = head_ref;
head_ref = new_node;
return head_ref;
}
// Function to print nodes in
// a given linked list
function printList(node)
{
while (node != null)
{
document.write(node.data + " ");
node = node.next;
}
}
// Driver code
var start = null;
// The constructed linked list is:
// 1.2.3.4.5
start = push(start, 5);
start = push(start, 4);
start = push(start, 3);
start = push(start, 2);
start = push(start, 1);
document.write("Linked list before " +
"moving first to end<br>");
printList(start);
start = moveToEnd(start);
document.write("<br> Linked list after moving " +
"first to end<br>");
printList(start);
// This code is contributed by rdtank
</script>
Producción:
Linked list before moving first to end 1 2 3 4 5 Linked list after moving first to end 2 3 4 5 1
Complejidad de tiempo: O(n) donde n es el número de Nodes en la lista enlazada dada.
Publicación traducida automáticamente
Artículo escrito por aganjali10 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA