Dadas dos listas ordenadas en orden creciente, cree y devuelva una nueva lista que represente la intersección de las dos listas. La nueva lista debe hacerse con su propia memoria; las listas originales no deben cambiarse.
Ejemplo:
Input: First linked list: 1->2->3->4->6 Second linked list be 2->4->6->8, Output: 2->4->6. The elements 2, 4, 6 are common in both the list so they appear in the intersection list. Input: First linked list: 1->2->3->4->5 Second linked list be 2->3->4, Output: 2->3->4 The elements 2, 3, 4 are common in both the list so they appear in the intersection list.
Método 1 : Uso del Node ficticio.
Enfoque:
la idea es utilizar un Node ficticio temporal al comienzo de la lista de resultados. La cola del puntero siempre apunta al último Node en la lista de resultados, por lo que se pueden agregar nuevos Nodes fácilmente. El Node ficticio inicialmente le da a la cola un espacio de memoria al que apuntar. Este Node ficticio es eficiente, ya que solo es temporal y se asigna en la pila. El ciclo continúa, eliminando un Node de ‘a’ o ‘b’ y agregándolo a la cola. Cuando se recorren las listas dadas, el resultado es ficticio. siguiente, ya que los valores se asignan desde el siguiente Node del dummy. Si ambos elementos son iguales, elimine ambos e inserte el elemento en la cola. De lo contrario, elimine el elemento más pequeño entre ambas listas.
A continuación se muestra la implementación del enfoque anterior:
C++
#include<bits/stdc++.h>
using namespace std;
/* Link list node */
struct Node {
int data;
Node* next;
};
void push(Node** head_ref, int new_data);
/*This solution uses the temporary
dummy to build up the result list */
Node* sortedIntersect(Node* a, Node* b)
{
Node dummy;
Node* tail = &dummy;
dummy.next = NULL;
/* Once one or the other
list runs out -- we're done */
while (a != NULL && b != NULL) {
if (a->data == b->data) {
push((&tail->next), a->data);
tail = tail->next;
a = a->next;
b = b->next;
}
/* advance the smaller list */
else if (a->data < b->data)
a = a->next;
else
b = b->next;
}
return (dummy.next);
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at
the beginning of the linked list */
void push(Node** head_ref, int new_data)
{
/* allocate node */
Node* new_node = (Node*)malloc(
sizeof(Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print nodes in
a given linked list */
void printList(Node* node)
{
while (node != NULL) {
cout << node->data <<" ";
node = node->next;
}
}
/* Driver program to test above functions*/
int main()
{
/* Start with the empty lists */
Node* a = NULL;
Node* b = NULL;
Node* intersect = NULL;
/* Let us create the first sorted
linked list to test the functions
Created linked list will be
1->2->3->4->5->6 */
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
/* Let us create the second sorted linked list
Created linked list will be 2->4->6->8 */
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
cout<<"Linked list containing common items of a & b \n";
printList(intersect);
}
C
#include <stdio.h>
#include <stdlib.h>
/* Link list node */
struct Node {
int data;
struct Node* next;
};
void push(struct Node** head_ref, int new_data);
/*This solution uses the temporary
dummy to build up the result list */
struct Node* sortedIntersect(
struct Node* a,
struct Node* b)
{
struct Node dummy;
struct Node* tail = &dummy;
dummy.next = NULL;
/* Once one or the other
list runs out -- we're done */
while (a != NULL && b != NULL) {
if (a->data == b->data) {
push((&tail->next), a->data);
tail = tail->next;
a = a->next;
b = b->next;
}
/* advance the smaller list */
else if (a->data < b->data)
a = a->next;
else
b = b->next;
}
return (dummy.next);
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at
the beginning of the linked list */
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node = (struct Node*)malloc(
sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*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 functions*/
int main()
{
/* Start with the empty lists */
struct Node* a = NULL;
struct Node* b = NULL;
struct Node* intersect = NULL;
/* Let us create the first sorted
linked list to test the functions
Created linked list will be
1->2->3->4->5->6 */
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
/* Let us create the second sorted linked list
Created linked list will be 2->4->6->8 */
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
printf("\n Linked list containing common items of a & b \n ");
printList(intersect);
getchar();
}
Java
class GFG
{
// head nodes for pointing to 1st and 2nd linked lists
static Node a = null, b = null;
// dummy node for storing intersection
static Node dummy = null;
// tail node for keeping track of
// last node so that it makes easy for insertion
static Node tail = null;
// class - Node
static class Node {
int data;
Node next;
Node(int data) {
this.data = data;
next = null;
}
}
// function for printing the list
void printList(Node start) {
Node p = start;
while (p != null) {
System.out.print(p.data + " ");
p = p.next;
}
System.out.println();
}
// inserting elements into list
void push(int data) {
Node temp = new Node(data);
if(dummy == null) {
dummy = temp;
tail = temp;
}
else {
tail.next = temp;
tail = temp;
}
}
// function for finding intersection and adding it to dummy list
void sortedIntersect()
{
// pointers for iterating
Node p = a,q = b;
while(p != null && q != null)
{
if(p.data == q.data)
{
// add to dummy list
push(p.data);
p = p.next;
q = q.next;
}
else if(p.data < q.data)
p = p.next;
else
q= q.next;
}
}
// Driver code
public static void main(String args[])
{
GFG list = new GFG();
// creating first linked list
list.a = new Node(1);
list.a.next = new Node(2);
list.a.next.next = new Node(3);
list.a.next.next.next = new Node(4);
list.a.next.next.next.next = new Node(6);
// creating second linked list
list.b = new Node(2);
list.b.next = new Node(4);
list.b.next.next = new Node(6);
list.b.next.next.next = new Node(8);
// function call for intersection
list.sortedIntersect();
// print required intersection
System.out.println("Linked list containing common items of a & b");
list.printList(dummy);
}
}
// This code is contributed by Likhita AVL
Python3
''' Link list node '''
class Node:
def __init__(self):
self.data = 0
self.next = None
'''This solution uses the temporary
dummy to build up the result list '''
def sortedIntersect(a, b):
dummy = Node()
tail = dummy;
dummy.next = None;
''' Once one or the other
list runs out -- we're done '''
while (a != None and b != None):
if (a.data == b.data):
tail.next = push((tail.next), a.data);
tail = tail.next;
a = a.next;
b = b.next;
# advance the smaller list
elif(a.data < b.data):
a = a.next;
else:
b = b.next;
return (dummy.next);
''' UTILITY FUNCTIONS '''
''' Function to insert a node at
the beginning of the linked list '''
def push(head_ref, new_data):
''' allocate node '''
new_node = Node()
''' put in the data '''
new_node.data = new_data;
''' link the old list off the new node '''
new_node.next = (head_ref);
''' move the head to point to the new node '''
(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'''
if __name__=='__main__':
''' Start with the empty lists '''
a = None;
b = None;
intersect = None;
''' Let us create the first sorted
linked list to test the functions
Created linked list will be
1.2.3.4.5.6 '''
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
''' Let us create the second sorted linked list
Created linked list will be 2.4.6.8 '''
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
''' Find the intersection two linked lists '''
intersect = sortedIntersect(a, b);
print("Linked list containing common items of a & b ");
printList(intersect);
# This code is contributed by rutvik_56.
C#
using System;
public class GFG
{
// dummy node for storing intersection
static Node dummy = null;
// tail node for keeping track of
// last node so that it makes easy for insertion
static Node tail = null;
// class - Node
public class Node
{
public int data;
public Node next;
public Node(int data)
{
this.data = data;
next = null;
}
}
// head nodes for pointing to 1st and 2nd linked lists
Node a = null, b = null;
// function for printing the list
void printList(Node start)
{
Node p = start;
while (p != null)
{
Console.Write(p.data + " ");
p = p.next;
}
Console.WriteLine();
}
// inserting elements into list
void push(int data)
{
Node temp = new Node(data);
if(dummy == null)
{
dummy = temp;
tail = temp;
}
else
{
tail.next = temp;
tail = temp;
}
}
// function for finding intersection and adding it to dummy list
void sortedIntersect()
{
// pointers for iterating
Node p = a,q = b;
while(p != null && q != null)
{
if(p.data == q.data)
{
// add to dummy list
push(p.data);
p = p.next;
q = q.next;
}
else if(p.data < q.data)
p = p.next;
else
q= q.next;
}
}
// Driver code
public static void Main(String []args)
{
GFG list = new GFG();
// creating first linked list
list.a = new Node(1);
list.a.next = new Node(2);
list.a.next.next = new Node(3);
list.a.next.next.next = new Node(4);
list.a.next.next.next.next = new Node(6);
// creating second linked list
list.b = new Node(2);
list.b.next = new Node(4);
list.b.next.next = new Node(6);
list.b.next.next.next = new Node(8);
// function call for intersection
list.sortedIntersect();
// print required intersection
Console.WriteLine("Linked list containing common items of a & b");
list.printList(dummy);
}
}
// This code is contributed by aashish1995
Javascript
<script>
// head nodes for pointing to
// 1st and 2nd linked lists
var a = null, b = null;
// dummy node for storing intersection
var dummy = null;
// tail node for keeping track of
// last node so that it makes easy for insertion
var tail = null;
// class - Node
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
// function for printing the list
function printList(start) {
var p = start;
while (p != null) {
document.write(p.data + " ");
p = p.next;
}
document.write();
}
// inserting elements into list
function push(data) {
var temp = new Node(data);
if (dummy == null) {
dummy = temp;
tail = temp;
} else {
tail.next = temp;
tail = temp;
}
}
// function for finding intersection and
// adding it to dummy list
function sortedIntersect() {
// pointers for iterating
var p = a, q = b;
while (p != null && q != null) {
if (p.data == q.data) {
// add to dummy list
push(p.data);
p = p.next;
q = q.next;
} else if (p.data < q.data)
p = p.next;
else
q = q.next;
}
}
// Driver code
// creating first linked list
a = new Node(1);
a.next = new Node(2);
a.next.next = new Node(3);
a.next.next.next = new Node(4);
a.next.next.next.next = new Node(6);
// creating second linked list
b = new Node(2);
b.next = new Node(4);
b.next.next = new Node(6);
b.next.next.next = new Node(8);
// function call for intersection
sortedIntersect();
// print required intersection
document.write(
"Linked list containing common items of a & b<br/>"
);
printList(dummy);
// This code is contributed by todaysgaurav
</script>
Producción
Linked list containing common items of a & b 2 4 6
Análisis de Complejidad:
- Complejidad de tiempo: O(m+n) donde m y n son el número de Nodes en la primera y segunda lista enlazada respectivamente.
Solo se necesita un recorrido de las listas. - Espacio Auxiliar: O(min(m, n)).
La lista de salida puede almacenar como máximo min(m,n) Nodes.
Método 2 : uso de referencias locales.
Enfoque: esta solución es estructuralmente muy similar a la anterior, pero evita usar un Node ficticio. En su lugar, mantiene un puntero de struct node**, lastPtrRef, que siempre apunta al último puntero de la lista de resultados. Esto resuelve el mismo caso que solucionó el Node ficticio: lidiar con la lista de resultados cuando está vacía. Si la lista se construye en su cola, se puede usar el Node ficticio o la estrategia de «referencia» de struct node**.
A continuación se muestra la implementación del enfoque anterior:
C++14
// C++ program to implement above approach
#include <bits/stdc++.h>
/* Link list node */
struct Node {
int data;
struct Node* next;
};
void push(struct Node** head_ref,
int new_data);
/* This solution uses the local reference */
struct Node* sortedIntersect(
struct Node* a,
struct Node* b)
{
struct Node* result = NULL;
struct Node** lastPtrRef = &result;
/* Advance comparing the first
nodes in both lists.
When one or the other list runs
out, we're done. */
while (a != NULL && b != NULL) {
if (a->data == b->data) {
/* found a node for the intersection */
push(lastPtrRef, a->data);
lastPtrRef = &((*lastPtrRef)->next);
a = a->next;
b = b->next;
}
else if (a->data < b->data)
a = a->next; /* advance the smaller list */
else
b = b->next;
}
return (result);
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the
beginning of the linked list */
void push(struct Node** head_ref,
int new_data)
{
/* allocate node */
struct Node* new_node = (struct Node*)malloc(
sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*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 functions*/
int main()
{
/* Start with the empty lists */
struct Node* a = NULL;
struct Node* b = NULL;
struct Node* intersect = NULL;
/* Let us create the first sorted
linked list to test the functions
Created linked list will be
1->2->3->4->5->6 */
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
/* Let us create the second sorted linked list
Created linked list will be 2->4->6->8 */
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
printf("\n Linked list containing common items of a & b \n ");
printList(intersect);
return 0;
}
//This code is contributed by Abhijeet Kumar(abhijeet19403)
C
#include <stdio.h>
#include <stdlib.h>
/* Link list node */
struct Node {
int data;
struct Node* next;
};
void push(struct Node** head_ref,
int new_data);
/* This solution uses the local reference */
struct Node* sortedIntersect(
struct Node* a,
struct Node* b)
{
struct Node* result = NULL;
struct Node** lastPtrRef = &result;
/* Advance comparing the first
nodes in both lists.
When one or the other list runs
out, we're done. */
while (a != NULL && b != NULL) {
if (a->data == b->data) {
/* found a node for the intersection */
push(lastPtrRef, a->data);
lastPtrRef = &((*lastPtrRef)->next);
a = a->next;
b = b->next;
}
else if (a->data < b->data)
a = a->next; /* advance the smaller list */
else
b = b->next;
}
return (result);
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the
beginning of the linked list */
void push(struct Node** head_ref,
int new_data)
{
/* allocate node */
struct Node* new_node = (struct Node*)malloc(
sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*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 functions*/
int main()
{
/* Start with the empty lists */
struct Node* a = NULL;
struct Node* b = NULL;
struct Node* intersect = NULL;
/* Let us create the first sorted
linked list to test the functions
Created linked list will be
1->2->3->4->5->6 */
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
/* Let us create the second sorted linked list
Created linked list will be 2->4->6->8 */
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
printf("\n Linked list containing common items of a & b \n ");
printList(intersect);
getchar();
}
Java
// Java program to implement above approach
import java.util.*;
class GFG {
// Link list node
static class Node {
int data;
Node next;
Node(int d)
{
data = d;
next = null;
}
};
static Node sortedIntersect(Node a, Node b)
{
Node result = new Node(0);
Node curr = result;
/* Advance comparing the first
nodes in both lists.
When one or the other list runs
out, we're done. */
while (a != null && b != null) {
if (a.data == b.data) {
/* found a node for the intersection */
curr.next = new Node(a.data);
curr = curr.next;
a = a.next;
b = b.next;
}
else if (a.data < b.data)
a = a.next; /* advance the smaller list */
else
b = b.next;
}
result = result.next;
return result;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at
the beginning of the linked list */
static Node push(Node head_ref, int new_data)
{
/* Allocate node */
Node new_node = new Node(new_data);
/* Link the old list off the new node */
new_node.next = head_ref;
/* Move the head to point to the new node */
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.print(node.data + " ");
node = node.next;
}
}
// Driver code
public static void main(String[] args)
{
/* Start with the empty lists */
Node a = null;
Node b = null;
Node intersect = null;
/* Let us create the first sorted
linked list to test the functions
Created linked list will be
1.2.3.4.5.6 */
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
/* Let us create the second sorted linked list
Created linked list will be 2.4.6.8 */
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
System.out.print("\n Linked list containing "
+ "common items of a & b \n ");
printList(intersect);
}
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)
Python3
# Python3 program to implement above approach
# Link list node
class Node:
def __init__(self, d):
self.data = d
self.next = None
def sortedIntersect(a, b):
result = Node(0)
curr = result
# Advance comparing the first
# nodes in both lists.
# When one or the other list runs
# out, we're done.
while (a != None and b != None):
if (a.data == b.data):
# found a node for the intersection
curr.next = Node(a.data)
curr = curr.next
a = a.next
b = b.next
elif (a.data < b.data):
a = a.next # advance the smaller list
else:
b = b.next
result = result.next
return result
# UTILITY FUNCTIONS
# Function to insert a node at the beginning of the linked list
def push(head_ref, new_data):
# Allocate node
new_node = Node(new_data)
# Link the old list off the new node
new_node.next = head_ref
# Move the head to point to the new node
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 with the empty lists
a = None
b = None
intersect = None
# Let us create the first sorted linked list to test the functions Created
# linked list will be 1.2.3.4.5.6
a = push(a, 6)
a = push(a, 5)
a = push(a, 4)
a = push(a, 3)
a = push(a, 2)
a = push(a, 1)
# Let us create the second sorted linked list Created linked list will be
# 2.4.6.8
b = push(b, 8)
b = push(b, 6)
b = push(b, 4)
b = push(b, 2)
# Find the intersection two linked lists
intersect = sortedIntersect(a, b)
print("Linked list containing " + "common items of a & b")
printList(intersect)
# This code is contributed by Abhijeet Kumar(abhijeet19403)
C#
// C# program to implement above approach
using System;
public class GFG {
// Link list node
public class Node {
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
};
static Node sortedIntersect(Node a, Node b)
{
Node result = new Node(0);
Node curr = result;
/* Advance comparing the first
nodes in both lists.
When one or the other list runs
out, we're done. */
while (a != null && b != null) {
if (a.data == b.data) {
/* found a node for the intersection */
curr.next = new Node(a.data);
curr = curr.next;
a = a.next;
b = b.next;
}
else if (a.data < b.data)
a = a.next; /* advance the smaller list */
else
b = b.next;
}
result = result.next;
return result;
}
/* UTILITY FUNCTIONS */
/*
* Function to insert a node at the beginning of the
* linked list
*/
static Node push(Node head_ref, int new_data)
{
/* Allocate node */
Node new_node = new Node(new_data);
/* Link the old list off the new node */
new_node.next = head_ref;
/* Move the head to point to the new node */
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(node.data + " ");
node = node.next;
}
}
// Driver code
public static void Main(String[] args)
{
/* Start with the empty lists */
Node a = null;
Node b = null;
Node intersect = null;
/*
* Let us create the first sorted linked list to
* test the functions Created linked list will
* be 1.2.3.4.5.6
*/
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
/*
* Let us create the second sorted linked list
* Created linked list will be 2.4.6.8
*/
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
Console.Write("\n Linked list containing "
+ "common items of a & b \n ");
printList(intersect);
}
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)
Producción
Linked list containing common items of a & b 2 4 6
Análisis de Complejidad:
- Complejidad de tiempo: O(m+n) donde m y n son el número de Nodes en la primera y segunda lista enlazada respectivamente.
Solo se necesita un recorrido de las listas. - Espacio Auxiliar: O(max(m, n)).
La lista de salida puede almacenar como máximo m+n Nodes.
Método 3 : Solución recursiva.
Enfoque:
El enfoque recursivo es muy similar a los dos enfoques anteriores. Cree una función recursiva que tome dos Nodes y devuelva un Node de lista enlazada. Compare el primer elemento de ambas listas.
- Si son similares, llame a la función recursiva con el siguiente Node de ambas listas. Cree un Node con los datos del Node actual y coloque el Node devuelto por la función recursiva en el siguiente puntero del Node creado. Devuelve el Node creado.
- Si los valores no son iguales, elimine el Node más pequeño de ambas listas y llame a la función recursiva.
A continuación se muestra la implementación del enfoque anterior:
C++
#include <bits/stdc++.h>
using namespace std;
// Link list node
struct Node
{
int data;
struct Node* next;
};
struct Node* sortedIntersect(struct Node* a,
struct Node* b)
{
// base case
if (a == NULL || b == NULL)
return NULL;
/* If both lists are non-empty */
/* Advance the smaller list and
call recursively */
if (a->data < b->data)
return sortedIntersect(a->next, b);
if (a->data > b->data)
return sortedIntersect(a, b->next);
// Below lines are executed only
// when a->data == b->data
struct Node* temp = (struct Node*)malloc(
sizeof(struct Node));
temp->data = a->data;
// Advance both lists and call recursively
temp->next = sortedIntersect(a->next,
b->next);
return temp;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at
the beginning of the linked list */
void push(struct Node** head_ref, int new_data)
{
/* Allocate node */
struct Node* new_node = (struct Node*)malloc(
sizeof(struct Node));
/* Put in the data */
new_node->data = new_data;
/* Link the old list off the new node */
new_node->next = (*head_ref);
/* Move the head to point to the new node */
(*head_ref) = new_node;
}
/* Function to print nodes in
a given linked list */
void printList(struct Node* node)
{
while (node != NULL)
{
cout << " " << node->data;
node = node->next;
}
}
// Driver code
int main()
{
/* Start with the empty lists */
struct Node* a = NULL;
struct Node* b = NULL;
struct Node* intersect = NULL;
/* Let us create the first sorted
linked list to test the functions
Created linked list will be
1->2->3->4->5->6 */
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
/* Let us create the second sorted linked list
Created linked list will be 2->4->6->8 */
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
cout << "\n Linked list containing "
<< "common items of a & b \n ";
printList(intersect);
return 0;
}
// This code is contributed by shivanisinghss2110
C
#include <stdio.h>
#include <stdlib.h>
/* Link list node */
struct Node {
int data;
struct Node* next;
};
struct Node* sortedIntersect(
struct Node* a,
struct Node* b)
{
/* base case */
if (a == NULL || b == NULL)
return NULL;
/* If both lists are non-empty */
/* advance the smaller list and call recursively */
if (a->data < b->data)
return sortedIntersect(a->next, b);
if (a->data > b->data)
return sortedIntersect(a, b->next);
// Below lines are executed only
// when a->data == b->data
struct Node* temp
= (struct Node*)malloc(
sizeof(struct Node));
temp->data = a->data;
/* advance both lists and call recursively */
temp->next = sortedIntersect(a->next, b->next);
return temp;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at
the beginning of the linked list */
void push(struct Node** head_ref, int new_data)
{
/* allocate node */
struct Node* new_node
= (struct Node*)malloc(
sizeof(struct Node));
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*head_ref);
/* move the head to point to the new node */
(*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 functions*/
int main()
{
/* Start with the empty lists */
struct Node* a = NULL;
struct Node* b = NULL;
struct Node* intersect = NULL;
/* Let us create the first sorted
linked list to test the functions
Created linked list will be
1->2->3->4->5->6 */
push(&a, 6);
push(&a, 5);
push(&a, 4);
push(&a, 3);
push(&a, 2);
push(&a, 1);
/* Let us create the second sorted linked list
Created linked list will be 2->4->6->8 */
push(&b, 8);
push(&b, 6);
push(&b, 4);
push(&b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
printf("\n Linked list containing common items of a & b \n ");
printList(intersect);
return 0;
}
Java
import java.util.*;
class GFG{
// Link list node
static class Node
{
int data;
Node next;
};
static Node sortedIntersect(Node a,
Node b)
{
// base case
if (a == null || b == null)
return null;
/* If both lists are non-empty */
/* Advance the smaller list and
call recursively */
if (a.data < b.data)
return sortedIntersect(a.next, b);
if (a.data > b.data)
return sortedIntersect(a, b.next);
// Below lines are executed only
// when a.data == b.data
Node temp = new Node();
temp.data = a.data;
// Advance both lists and call recursively
temp.next = sortedIntersect(a.next,
b.next);
return temp;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at
the beginning of the linked list */
static Node push(Node head_ref, int new_data)
{
/* Allocate node */
Node new_node = new Node();
/* Put in the data */
new_node.data = new_data;
/* Link the old list off the new node */
new_node.next = head_ref;
/* Move the head to point to the new node */
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.print(" " + node.data);
node = node.next;
}
}
// Driver code
public static void main(String[] args)
{
/* Start with the empty lists */
Node a = null;
Node b = null;
Node intersect = null;
/* Let us create the first sorted
linked list to test the functions
Created linked list will be
1.2.3.4.5.6 */
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
/* Let us create the second sorted linked list
Created linked list will be 2.4.6.8 */
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
System.out.print("\n Linked list containing "
+ "common items of a & b \n ");
printList(intersect);
}
}
// This code is contributed by umadevi9616
Python3
# Link list node
class Node:
def __init__(self):
self.data = 0
self.next = None
def sortedIntersect(a, b):
# base case
if (a == None or b == None):
return None
# If both lists are non-empty
# Advance the smaller list and call recursively
if (a.data < b.data):
return sortedIntersect(a.next, b);
if (a.data > b.data):
return sortedIntersect(a, b.next);
# Below lines are executed only
# when a.data == b.data
temp = Node();
temp.data = a.data;
# Advance both lists and call recursively
temp.next = sortedIntersect(a.next, b.next);
return temp;
# UTILITY FUNCTIONS
# Function to insert a node at the beginning of the linked list
def push(head_ref, new_data):
# Allocate node
new_node = Node()
# Put in the data
new_node.data = new_data;
# Link the old list off the new node
new_node.next = head_ref;
# Move the head to point to the new node
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 with the empty lists
a = None
b = None
intersect = None
# Let us create the first sorted linked list to test the functions Created
# linked list will be 1.2.3.4.5.6
a = push(a, 6)
a = push(a, 5)
a = push(a, 4)
a = push(a, 3)
a = push(a, 2)
a = push(a, 1)
# Let us create the second sorted linked list Created linked list will be
# 2.4.6.8
b = push(b, 8)
b = push(b, 6)
b = push(b, 4)
b = push(b, 2)
# Find the intersection two linked lists
intersect = sortedIntersect(a, b)
print("\n Linked list containing " + "common items of a & b");
printList(intersect)
# This code is contributed by Saurabh Jaiswal
C#
using System;
public class GFG {
// Link list node
public class Node {
public int data;
public Node next;
};
static Node sortedIntersect(Node a, Node b) {
// base case
if (a == null || b == null)
return null;
/* If both lists are non-empty */
/*
* Advance the smaller list and call recursively
*/
if (a.data < b.data)
return sortedIntersect(a.next, b);
if (a.data > b.data)
return sortedIntersect(a, b.next);
// Below lines are executed only
// when a.data == b.data
Node temp = new Node();
temp.data = a.data;
// Advance both lists and call recursively
temp.next = sortedIntersect(a.next, b.next);
return temp;
}
/* UTILITY FUNCTIONS */
/*
* Function to insert a node at the beginning of the linked list
*/
static Node push(Node head_ref, int new_data) {
/* Allocate node */
Node new_node = new Node();
/* Put in the data */
new_node.data = new_data;
/* Link the old list off the new node */
new_node.next = head_ref;
/* Move the head to point to the new node */
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(" " + node.data);
node = node.next;
}
}
// Driver code
public static void Main(String[] args) {
/* Start with the empty lists */
Node a = null;
Node b = null;
Node intersect = null;
/*
* Let us create the first sorted linked list to test the functions Created
* linked list will be 1.2.3.4.5.6
*/
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
/*
* Let us create the second sorted linked list Created linked list will be
* 2.4.6.8
*/
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
Console.Write("\n Linked list containing " + "common items of a & b \n ");
printList(intersect);
}
}
// This code is contributed by umadevi9616
Javascript
<script>
// Link list node
class Node {
constructor(){
this.data = 0;
this.next = null;
}
}
function sortedIntersect(a, b) {
// base case
if (a == null || b == null)
return null;
/* If both lists are non-empty */
/*
* Advance the smaller list and call recursively
*/
if (a.data < b.data)
return sortedIntersect(a.next, b);
if (a.data > b.data)
return sortedIntersect(a, b.next);
// Below lines are executed only
// when a.data == b.data
var temp = new Node();
temp.data = a.data;
// Advance both lists and call recursively
temp.next = sortedIntersect(a.next, b.next);
return temp;
}
/* UTILITY FUNCTIONS */
/*
* Function to insert a node at the beginning of the linked list
*/
function push(head_ref , new_data) {
/* Allocate node */
var new_node = new Node();
/* Put in the data */
new_node.data = new_data;
/* Link the old list off the new node */
new_node.next = head_ref;
/* Move the head to point to the new node */
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
/* Start with the empty lists */
var a = null;
var b = null;
var intersect = null;
/*
* Let us create the first sorted linked list to test the functions Created
* linked list will be 1.2.3.4.5.6
*/
a = push(a, 6);
a = push(a, 5);
a = push(a, 4);
a = push(a, 3);
a = push(a, 2);
a = push(a, 1);
/*
* Let us create the second sorted linked list Created linked list will be
* 2.4.6.8
*/
b = push(b, 8);
b = push(b, 6);
b = push(b, 4);
b = push(b, 2);
/* Find the intersection two linked lists */
intersect = sortedIntersect(a, b);
document.write("\n Linked list containing "
+ "common items of a & b <br/> ");
printList(intersect);
// This code is contributed by Rajput-Ji
</script>
Producción
Linked list containing common items of a & b 2 4 6
Análisis de Complejidad:
- Complejidad de tiempo: O(m+n) donde m y n son el número de Nodes en la primera y segunda lista enlazada respectivamente.
Solo se necesita un recorrido de las listas. - Espacio Auxiliar: O(max(m, n)).
La lista de salida puede almacenar como máximo m+n Nodes.
Método 4: Usar hashing
C++14
// C++ program to implement above approach
#include <bits/stdc++.h>
using namespace std;
// Link list node
struct Node {
int data;
struct Node* next;
};
void printList(struct Node* node)
{
while (node != NULL) {
cout << " " << node->data;
node = node->next;
}
}
void append(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;
}
vector<int> intersection(struct Node* tmp1, struct Node* tmp2,
int k)
{
vector<int> res(k);
unordered_set<int> set;
while (tmp1 != NULL) {
set.insert(tmp1->data);
tmp1 = tmp1->next;
}
int cnt = 0;
while (tmp2 != NULL) {
if (set.find(tmp2->data) != set.end()) {
res[cnt] = tmp2->data;
cnt++;
}
tmp2 = tmp2->next;
}
return res;
}
// Driver code
int main()
{
struct Node* ll = NULL;
struct Node* ll1 = NULL;
append(&ll,7);
append(&ll,6);
append(&ll,5);
append(&ll,4);
append(&ll,3);
append(&ll,2);
append(&ll,1);
append(&ll,0);
append(&ll1,7);
append(&ll1,6);
append(&ll1,5);
append(&ll1,4);
append(&ll1,3);
append(&ll1,12);
append(&ll1,0);
append(&ll1,9);
vector<int> arr= intersection(ll, ll1, 6);
for (int i :arr)
cout << i << "\n";
return 0;
}
// This code is contributed by Abhijeet Kumar(abhijeet19403)
Java
import java.util.*;
// This code is contributed by ayyuce demirbas
public class LinkedList {
Node head;
static class Node {
int data;
Node next;
Node(int d) {
data = d;
next=null;
}
}
public void printList() {
Node n= head;
while(n!=null) {
System.out.println(n.data+ " ");
n=n.next;
}
}
public void append(int d) {
Node n= new Node(d);
if(head== null) {
head= new Node(d);
return;
}
n.next=null;
Node last= head;
while(last.next !=null) {
last=last.next;
}
last.next=n;
return;
}
static int[] intersection(Node tmp1, Node tmp2, int k) {
int[] res = new int[k];
HashSet<Integer> set = new HashSet<Integer>();
while(tmp1 != null) {
set.add(tmp1.data);
tmp1=tmp1.next;
}
int cnt=0;
while(tmp2 != null) {
if(set.contains(tmp2.data)) {
res[cnt]=tmp2.data;
cnt++;
}
tmp2=tmp2.next;
}
return res;
}
public static void main(String[] args) {
LinkedList ll = new LinkedList();
LinkedList ll1 = new LinkedList();
ll.append(0);
ll.append(1);
ll.append(2);
ll.append(3);
ll.append(4);
ll.append(5);
ll.append(6);
ll.append(7);
ll1.append(9);
ll1.append(0);
ll1.append(12);
ll1.append(3);
ll1.append(4);
ll1.append(5);
ll1.append(6);
ll1.append(7);
int[] arr= intersection(ll.head, ll1.head,6);
for(int i : arr) {
System.out.println(i);
}
}
}
Python3
# Python3 program to implement above approach # Link list node class Node: def __init__(self): self.data = 0 self.next = None def printList(node): while (node != None): print(node.data, end=" ") node = node.next; def append(head_ref, new_data): new_node = Node() new_node.data = new_data; new_node.next = head_ref; head_ref = new_node; return head_ref; def intersection(tmp1,tmp2,k): res = [0]*k set1 = set() while (tmp1 != None): set1.add(tmp1.data) tmp1 = tmp1.next cnt = 0 while (tmp2 != None): if tmp2.data in set1: res[cnt] = tmp2.data; cnt += 1 tmp2 = tmp2.next return res def printList(node): while (node != None): print(node.data, end=" ") node = node.next; # Driver code # Start with the empty lists ll = None ll1 = None ll = append(ll , 7) ll = append(ll , 6) ll = append(ll , 5) ll = append(ll , 4) ll = append(ll , 3) ll = append(ll , 2) ll = append(ll , 1) ll = append(ll , 0) ll1 = append(ll1 , 7) ll1 = append(ll1 , 6) ll1 = append(ll1 , 5) ll1 = append(ll1 , 4) ll1 = append(ll1 , 3) ll1 = append(ll1 , 12) ll1 = append(ll1 , 0) ll1 = append(ll1 , 9) arr = intersection(ll , ll1 , 6) for i in range(6): print(arr[i]) # This code is contributed by Abhijeet Kumar(abhijeet19403)
C#
using System;
using System.Collections.Generic;
// This code is contributed by ayyuce demirbas
public class List {
Node head;
public class Node {
public int data;
public Node next;
public Node(int d)
{
data = d;
next = null;
}
}
public void printList()
{
Node n = head;
while (n != null) {
Console.WriteLine(n.data + " ");
n = n.next;
}
}
public void append(int d)
{
Node n = new Node(d);
if (head == null) {
head = new Node(d);
return;
}
n.next = null;
Node last = head;
while (last.next != null) {
last = last.next;
}
last.next = n;
return;
}
static int[] intersection(Node tmp1, Node tmp2, int k)
{
int[] res = new int[k];
HashSet<int> set = new HashSet<int>();
while (tmp1 != null) {
set.Add(tmp1.data);
tmp1 = tmp1.next;
}
int cnt = 0;
while (tmp2 != null) {
if (set.Contains(tmp2.data)) {
res[cnt] = tmp2.data;
cnt++;
}
tmp2 = tmp2.next;
}
return res;
}
public static void Main(String[] args)
{
List ll = new List();
List ll1 = new List();
ll.append(0);
ll.append(1);
ll.append(2);
ll.append(3);
ll.append(4);
ll.append(5);
ll.append(6);
ll.append(7);
ll1.append(9);
ll1.append(0);
ll1.append(12);
ll1.append(3);
ll1.append(4);
ll1.append(5);
ll1.append(6);
ll1.append(7);
int[] arr = intersection(ll.head, ll1.head, 6);
foreach(int i in arr) { Console.WriteLine(i); }
}
}
// This code is contributed by umadevi9616
Producción
0 3 4 5 6 7
Análisis de Complejidad:
- Complejidad de tiempo: O(n)
- Complejidad del espacio: O(n) ya que se está utilizando el espacio auxiliar
Escriba comentarios si encuentra que los códigos/algoritmos anteriores son incorrectos o encuentra mejores formas de resolver el mismo problema.
Referencias:
cslibrary.stanford.edu/105/LinkedListProblems.pdf
Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA