Dada una lista enlazada individualmente, elimine todos los Nodes que tienen un valor mayor en el lado derecho.
Ejemplos:
Input: 12->15->10->11->5->6->2->3->NULL
Output: 15->11->6->3->NULL
Explanation: 12, 10, 5 and 2 have been deleted because there is a
greater value on the right side. When we examine 12,
we see that after 12 there is one node with a value
greater than 12 (i.e. 15), so we delete 12. When we
examine 15, we find no node after 15 that has a value
greater than 15, so we keep this node. When we go like
this, we get 15->6->3
Input: 10->20->30->40->50->60->NULL
Output: 60->NULL
Explanation: 10, 20, 30, 40, and 50 have been deleted because
they all have a greater value on the right side.
Input: 60->50->40->30->20->10->NULL
Output: No Change.
Método 1 (Simple):
Use dos bucles. En el bucle exterior, seleccione los Nodes de la lista vinculada uno por uno. En el bucle interno, verifique si existe un Node cuyo valor sea mayor que el Node seleccionado. Si existe un Node cuyo valor es mayor, elimine el Node elegido.
Complejidad del tiempo: O(n^2)
Método 2 (Uso inverso):
gracias a Paras por proporcionar el siguiente algoritmo.
1. Invierta la lista.
2. Recorra la lista invertida. Mantenga el máximo hasta ahora. Si el siguiente Node es menor que max, elimine el siguiente Node; de lo contrario, max = next node.
3. Invierta la lista nuevamente para conservar el orden original.
Complejidad de tiempo: O(n)
Gracias a R.Srinivasan por proporcionar el código a continuación.
C
// C program to delete nodes which have
// a greater value on right side
#include <stdio.h>
#include <stdlib.h>
// Structure of a linked list
// node
struct Node
{
int data;
struct Node* next;
};
// Prototype for utility functions
void reverseList(struct Node** headref);
void _delLesserNodes(struct Node* head);
/* Deletes nodes which have a node with
greater value node on left side */
void delLesserNodes(struct Node** head_ref)
{
// 1. Reverse the linked list
reverseList(head_ref);
/* 2. In the reversed list, delete nodes
which have a node with greater value
node on left side. Note that head node
is never deleted because it is the leftmost
node.*/
_delLesserNodes(*head_ref);
/* 3. Reverse the linked list again to retain the
original order */
reverseList(head_ref);
}
/* Deletes nodes which have greater value
node(s) on left side */
void _delLesserNodes(struct Node* head)
{
struct Node* current = head;
// Initialize max
struct Node* maxnode = head;
struct Node* temp;
while (current != NULL &&
current->next != NULL)
{
/* If current is smaller than max,
then delete current */
if (current->next->data <
maxnode->data)
{
temp = current->next;
current->next = temp->next;
free(temp);
}
/* If current is greater than max,
then update max and move current */
else
{
current = current->next;
maxnode = current;
}
}
}
/* Utility function to insert a
node at the beginning */
void push(struct Node** head_ref,
int new_data)
{
struct Node* new_node =
(struct Node*)malloc(sizeof(struct Node));
new_node->data = new_data;
new_node->next = *head_ref;
*head_ref = new_node;
}
/* Utility function to reverse a
linked list */
void reverseList(struct Node** headref)
{
struct Node* current = *headref;
struct Node* prev = NULL;
struct Node* next;
while (current != NULL)
{
next = current->next;
current->next = prev;
prev = current;
current = next;
}
*headref = prev;
}
/* Utility function to print a
linked list */
void printList(struct Node* head)
{
while (head != NULL)
{
printf("%d ", head->data);
head = head->next;
}
printf("");
}
// Driver code
int main()
{
struct Node* head = NULL;
/* Create following linked list
12->15->10->11->5->6->2->3 */
push(&head, 3);
push(&head, 2);
push(&head, 6);
push(&head, 5);
push(&head, 11);
push(&head, 10);
push(&head, 15);
push(&head, 12);
printf("Given Linked List ");
printList(head);
delLesserNodes(&head);
printf("Modified Linked List ");
printList(head);
return 0;
}
Producción:
Given Linked List 12 15 10 11 5 6 2 3 Modified Linked List 15 11 6 3
Complejidad de tiempo : O(n)
Espacio Auxiliar : O(1)
Consulte el artículo completo sobre Eliminar Nodes que tienen un valor mayor en el lado derecho para obtener más detalles.
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA