Dada una lista de enlaces simples 
que está ordenada en orden ascendente y otra lista de enlaces simples 
que no está ordenada. La tarea es imprimir los elementos de la segunda lista enlazada según la posición señalada por los datos en los Nodes de la primera lista enlazada. Por ejemplo, si la primera lista enlazada es 1->2->5 , entonces debe imprimir los elementos 1 , 2 y 5 de la segunda lista enlazada .
Ejemplos :
Input: l1 = 1->2->5
l2 = 1->8->7->6->2->9->10
Output : 1->8->2
Elements in l1 are 1, 2 and 5.
Therefore, print 1st, 2nd and 5th elements of l2,
Which are 1, 8 and 2.
Input: l1 = 2->5
l2 = 7->5->3->2->8
Output: 5->8
Tome dos punteros para recorrer las dos listas vinculadas utilizando dos bucles anidados. El bucle exterior apunta a los elementos de la primera lista y el bucle interior apunta a los elementos de la segunda lista respectivamente. En la primera iteración del ciclo externo, el puntero al encabezado de la primera lista enlazada apunta a su Node raíz. Atravesamos la segunda lista enlazada hasta llegar a la posición señalada por los datos del Node en la primera lista enlazada. Una vez alcanzada la posición requerida, imprima los datos de la segunda lista y repita este proceso nuevamente hasta llegar al final de la primera lista enlazada.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program to print second linked list
// according to data in the first linked list
#include <iostream>
using namespace std;
// Linked List Node
struct Node {
int data;
struct Node* next;
};
/* Function to insert a node at the beginning */
void push(struct 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;
}
// Function to print the second list according
// to the positions referred by the 1st list
void printSecondList(Node* l1, Node* l2)
{
struct Node* temp = l1;
struct Node* temp1 = l2;
// While first linked list is not null.
while (temp != NULL) {
int i = 1;
// While second linked list is not equal
// to the data in the node of the
// first linked list.
while (i < temp->data) {
// Keep incrementing second list
temp1 = temp1->next;
i++;
}
// Print the node at position in second list
// pointed by current element of first list
cout << temp1->data << " ";
// Increment first linked list
temp = temp->next;
// Set temp1 to the start of the
// second linked list
temp1 = l2;
}
}
// Driver Code
int main()
{
struct Node *l1 = NULL, *l2 = NULL;
// Creating 1st list
// 2 -> 5
push(&l1, 5);
push(&l1, 2);
// Creating 2nd list
// 4 -> 5 -> 6 -> 7 -> 8
push(&l2, 8);
push(&l2, 7);
push(&l2, 6);
push(&l2, 5);
push(&l2, 4);
printSecondList(l1, l2);
return 0;
}
Java
// Java program to print second linked list
// according to data in the first linked list
class GFG
{
// Linked List Node
static class Node
{
int data;
Node next;
};
/* Function to insert a node at the beginning */
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 the second list according
// to the positions referred by the 1st list
static void printSecondList(Node l1, Node l2)
{
Node temp = l1;
Node temp1 = l2;
// While first linked list is not null.
while (temp != null)
{
int i = 1;
// While second linked list is not equal
// to the data in the node of the
// first linked list.
while (i < temp.data)
{
// Keep incrementing second list
temp1 = temp1.next;
i++;
}
// Print the node at position in second list
// pointed by current element of first list
System.out.print( temp1.data + " ");
// Increment first linked list
temp = temp.next;
// Set temp1 to the start of the
// second linked list
temp1 = l2;
}
}
// Driver Code
public static void main(String args[])
{
Node l1 = null, l2 = null;
// Creating 1st list
// 2 . 5
l1 = push(l1, 5);
l1 = push(l1, 2);
// Creating 2nd list
// 4 . 5 . 6 . 7 . 8
l2 = push(l2, 8);
l2 = push(l2, 7);
l2 = push(l2, 6);
l2 = push(l2, 5);
l2 = push(l2, 4);
printSecondList(l1, l2);
}
}
// This code is contributed by Arnab Kundu
Python3
# Python3 program to prsecond linked list # according to data in the first linked list # Linked List Node class new_Node: # Constructor to initialize the node object def __init__(self, data): self.data = data self.next = None ''' Function to insert a node at the beginning ''' def push(head_ref, new_data): new_node = new_Node(new_data) new_node.next = head_ref head_ref = new_node return head_ref # Function to print second list according # to the positions referred by the 1st list def printSecondList(l1,l2): temp = l1 temp1 = l2 # While first linked list is not None. while (temp != None): i = 1 # While second linked list is not equal # to the data in the node of the # first linked list. while (i < temp.data): # Keep incrementing second list temp1 = temp1.next i += 1 # Print node at position in second list # pointed by current element of first list print(temp1.data,end=" ") # Increment first linked list temp = temp.next # Set temp1 to the start of the # second linked list temp1 = l2 # Driver Code l1 = None l2 = None # Creating 1st list # 2 . 5 l1 = push(l1, 5) l1 = push(l1, 2) # Creating 2nd list # 4 . 5 . 6 . 7 . 8 l2 = push(l2, 8) l2 = push(l2, 7) l2 = push(l2, 6) l2 = push(l2, 5) l2 = push(l2, 4) printSecondList(l1, l2) # This code is contributed by shubhamsingh10
C#
// C# program to print second linked list
// according to data in the first linked list
using System;
class GFG
{
// Linked List Node
public class Node
{
public int data;
public Node next;
};
/* Function to insert a node at the beginning */
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 the second list according
// to the positions referred by the 1st list
static void printSecondList(Node l1, Node l2)
{
Node temp = l1;
Node temp1 = l2;
// While first linked list is not null.
while (temp != null)
{
int i = 1;
// While second linked list is not equal
// to the data in the node of the
// first linked list.
while (i < temp.data)
{
// Keep incrementing second list
temp1 = temp1.next;
i++;
}
// Print the node at position in second list
// pointed by current element of first list
Console.Write( temp1.data + " ");
// Increment first linked list
temp = temp.next;
// Set temp1 to the start of the
// second linked list
temp1 = l2;
}
}
// Driver Code
public static void Main()
{
Node l1 = null, l2 = null;
// Creating 1st list
// 2 . 5
l1 = push(l1, 5);
l1 = push(l1, 2);
// Creating 2nd list
// 4 . 5 . 6 . 7 . 8
l2 = push(l2, 8);
l2 = push(l2, 7);
l2 = push(l2, 6);
l2 = push(l2, 5);
l2 = push(l2, 4);
printSecondList(l1, l2);
}
}
// This code has been contributed by 29AjayKumar
Javascript
<script>
// JavaScript program to print second linked list
// according to data in the first linked list
class Node
{
constructor()
{
this.data=0;
this.next=null;
}
}
/* Function to insert a node at the beginning */
function push(head_ref, new_data)
{
let new_node = new Node();
new_node.data = new_data;
new_node.next = (head_ref);
(head_ref) = new_node;
return head_ref;
}
// Function to print the second list according
// to the positions referred by the 1st list
function printSecondList(l1,l2)
{
let temp = l1;
let temp1 = l2;
// While first linked list is not null.
while (temp != null)
{
let i = 1;
// While second linked list is not equal
// to the data in the node of the
// first linked list.
while (i < temp.data)
{
// Keep incrementing second list
temp1 = temp1.next;
i++;
}
// Print the node at position in second list
// pointed by current element of first list
document.write( temp1.data + " ");
// Increment first linked list
temp = temp.next;
// Set temp1 to the start of the
// second linked list
temp1 = l2;
}
}
// Driver Code
let l1 = null, l2 = null;
// Creating 1st list
// 2 . 5
l1 = push(l1, 5);
l1 = push(l1, 2);
// Creating 2nd list
// 4 . 5 . 6 . 7 . 8
l2 = push(l2, 8);
l2 = push(l2, 7);
l2 = push(l2, 6);
l2 = push(l2, 5);
l2 = push(l2, 4);
printSecondList(l1, l2);
// This code is contributed by avanitrachhadiya2155
</script>
Producción
5 8
Complejidad del tiempo: O(n^2)
Como estamos usando bucles anidados y recorriendo los elementos de la segunda lista para cada elemento de la primera lista.
Espacio Auxiliar: O(1)
Como espacio adicional constante se utiliza.
Otro enfoque:
Como sabemos que los datos de la primera lista están ordenados, podemos hacer uso de este hecho para recorrer la segunda lista.
- Haga dos punteros al encabezado de ambas listas.
- Poner un contador a 1
- Siga incrementando el contador y avanzando en la segunda lista hasta que el valor del contador sea menor que los datos en el puntero actual en la primera lista
- Imprima los datos del puntero de la segunda lista e incremente el puntero a la primera lista
- Repita los pasos 3 y 4 hasta que el puntero a la primera lista no sea NULL.
implementación:
C++
// C++ program to print second linked list
// according to data in the first linked list
#include <iostream>
using namespace std;
// Linked List Node
struct Node {
int data;
struct Node* next;
};
/* Function to insert a node at the beginning */
void push(struct 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;
}
// Function to print the second list according
// to the positions referred by the 1st list
void printSecondList(Node* l1, Node* l2)
{
struct Node* temp1 = l1;
struct Node* temp2 = l2;
int counter = 1;
// While first linked list is not null.
while (temp1 != NULL) {
// while the counter is less than the data at temp1
while (counter < temp1->data) {
// Keep incrementing second list
temp2 = temp2->next;
counter++;
}
// Print the node at position in second list
// pointed by current element of first list
cout << temp2->data << " ";
// Increment first linked list
temp1 = temp1->next;
}
}
// Driver Code
int main()
{
struct Node *l1 = NULL, *l2 = NULL;
// Creating 1st list
// 2 -> 5
push(&l1, 5);
push(&l1, 2);
// Creating 2nd list
// 4 -> 5 -> 6 -> 7 -> 8
push(&l2, 8);
push(&l2, 7);
push(&l2, 6);
push(&l2, 5);
push(&l2, 4);
printSecondList(l1, l2);
return 0;
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)
Java
// Java program to print second linked list
// according to data in the first linked list
class GFG {
// Linked List Node
static class Node {
int data;
Node next;
};
/* Function to insert a node at the beginning */
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 the second list according
// to the positions referred by the 1st list
static void printSecondList(Node l1, Node l2)
{
Node temp1 = l1;
Node temp2 = l2;
int counter = 1;
// While first linked list is not null.
while (temp1 != null) {
// while the counter is less than the data at
// temp1
while (counter < temp1.data) {
// Keep incrementing second list
temp2 = temp2.next;
counter++;
}
// Print the node at position in second list
// pointed by current element of first list
System.out.print(temp2.data + " ");
// Increment first linked list
temp1 = temp1.next;
}
}
// Driver Code
public static void main(String args[])
{
Node l1 = null, l2 = null;
// Creating 1st list
// 2 . 5
l1 = push(l1, 5);
l1 = push(l1, 2);
// Creating 2nd list
// 4 . 5 . 6 . 7 . 8
l2 = push(l2, 8);
l2 = push(l2, 7);
l2 = push(l2, 6);
l2 = push(l2, 5);
l2 = push(l2, 4);
printSecondList(l1, l2);
}
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)
Python3
# Python3 program to prsecond linked list # according to data in the first linked list # Linked List Node class new_Node: # Constructor to initialize the node object def __init__(self, data): self.data = data self.next = None ''' Function to insert a node at the beginning ''' def push(head_ref, new_data): new_node = new_Node(new_data) new_node.next = head_ref head_ref = new_node return head_ref # Function to print second list according # to the positions referred by the 1st list def printSecondList(l1, l2): temp1 = l1 temp2 = l2 counter = 1 # While first linked list is not None. while (temp1 != None): # while the counter is less than the data at temp1 while (counter < temp1.data): # Keep incrementing second list temp2 = temp2.next counter += 1 # Print node at position in second list # pointed by current element of first list print(temp2.data, end=" ") # Increment first linked list temp1 = temp1.next # Driver Code l1 = None l2 = None # Creating 1st list # 2 . 5 l1 = push(l1, 5) l1 = push(l1, 2) # Creating 2nd list # 4 . 5 . 6 . 7 . 8 l2 = push(l2, 8) l2 = push(l2, 7) l2 = push(l2, 6) l2 = push(l2, 5) l2 = push(l2, 4) printSecondList(l1, l2) # This code is contributed by Abhijeet Kumar(abhijeet19403)
C#
// C# program to print second linked list
// according to data in the first linked list
using System;
class GFG {
// Linked List Node
public class Node {
public int data;
public Node next;
};
/* Function to insert a node at the beginning */
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 the second list according
// to the positions referred by the 1st list
static void printSecondList(Node l1, Node l2)
{
Node temp1 = l1;
Node temp2 = l2;
int counter = 1;
// While first linked list is not null.
while (temp1 != null) {
// while the counter is less than the data at
// temp1
while (counter < temp1.data) {
// Keep incrementing second list
temp2 = temp2.next;
counter++;
}
// Print the node at position in second list
// pointed by current element of first list
Console.Write(temp2.data + " ");
// Increment first linked list
temp1 = temp1.next;
}
}
// Driver Code
public static void Main()
{
Node l1 = null, l2 = null;
// Creating 1st list
// 2 . 5
l1 = push(l1, 5);
l1 = push(l1, 2);
// Creating 2nd list
// 4 . 5 . 6 . 7 . 8
l2 = push(l2, 8);
l2 = push(l2, 7);
l2 = push(l2, 6);
l2 = push(l2, 5);
l2 = push(l2, 4);
printSecondList(l1, l2);
}
}
// This code has been contributed by Abhijeet Kumar(abhijeet19403)
Javascript
<script>
// JavaScript program to print second linked list
// according to data in the first linked list
class Node
{
constructor()
{
this.data=0;
this.next=null;
}
}
/* Function to insert a node at the beginning */
function push(head_ref, new_data)
{
let new_node = new Node();
new_node.data = new_data;
new_node.next = (head_ref);
(head_ref) = new_node;
return head_ref;
}
// Function to print the second list according
// to the positions referred by the 1st list
function printSecondList(l1,l2)
{
let temp1 = l1;
let temp2 = l2;
let counter = 1;
// While first linked list is not null.
while (temp1 != null)
{
// while the counter is less than the data at temp1
while (counter < temp1.data)
{
// Keep incrementing second list
temp2 = temp2.next;
counter++;
}
// Print the node at position in second list
// pointed by current element of first list
document.write( temp2.data + " ");
// Increment first linked list
temp1 = temp1.next;
}
}
// Driver Code
let l1 = null, l2 = null;
// Creating 1st list
// 2 . 5
l1 = push(l1, 5);
l1 = push(l1, 2);
// Creating 2nd list
// 4 . 5 . 6 . 7 . 8
l2 = push(l2, 8);
l2 = push(l2, 7);
l2 = push(l2, 6);
l2 = push(l2, 5);
l2 = push(l2, 4);
printSecondList(l1, l2);
// This code is contributed by Abhijeet Kumar(abhijeet19403)
</script>
Producción
5 8
Complejidad de tiempo: O(n)
Aunque estamos usando bucles anidados, el número total de operaciones está limitado solo a O (n).
Espacio Auxiliar: O(1)
Como espacio adicional constante se utiliza.
Este enfoque fue aportado por Abhijeet Kumar.
Escriba comentarios si encuentra algo incorrecto o si desea compartir más información sobre el tema tratado anteriormente.
Publicación traducida automáticamente
Artículo escrito por PRIYAJITDAS y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA