Dada una lista enlazada de un número par de Nodes, la tarea es generar una nueva lista enlazada que contenga la diferencia máxima de cuadrados de valores de Nodes en orden decreciente al incluir cada Node en un solo par.
Ejemplos:
Entrada: 1 -> 6 -> 4 -> 3 -> 5 ->2
Salida: 35 -> 21 -> 7
Explicación:
La diferencia entre cuadrados de 6 y 1 forma el primer Node con valor 35.
La diferencia entre cuadrados de 5 y 2 forma el segundo Node con valor 21.
La diferencia entre los cuadrados de 4 y 3 forma el tercer Node con valor 7.
Por lo tanto, la LL formada es 35 -> 21 -> 7.Entrada: 2 -> 4 -> 5 -> 3 -> 7 -> 8 -> 9 -> 10
Salida: 96 -> 72 -> 48 -> 10
Explicación:
La diferencia entre los cuadrados de 10 y 2 forma el primer Node con valor 96.
La diferencia entre cuadrados de 9 y 3 forma el segundo Node con valor 72.
La diferencia entre cuadrados de 8 y 4 forma el tercer Node con valor 48.
La diferencia entre cuadrados de 7 y 5 forma el cuarto Node con valor 10.
Por tanto, la LL formada es 96 -> 72 -> 48 -> 10.
Enfoque: El enfoque es encontrar el valor máximo de un Node y siempre hacer la diferencia entre el valor de Node más grande y el más pequeño. Así que cree un deque e inserte todos los valores de los Nodes en él, y ordene el deque . Ahora, acceda a los valores más grandes y más pequeños de ambos extremos. A continuación se muestran los pasos:
- Cree una deque e inserte todos los valores en la deque.
- Ordene el deque para obtener el valor de Node más grande y el valor de Node más pequeño en tiempo constante.
- Cree otra lista enlazada que tenga la diferencia de valores de los cuadrados de los valores más grandes y más pequeños de la parte posterior y frontal del deque, respectivamente.
- Después de cada iteración, saque tanto el valor más pequeño como el más grande del deque.
- Después de los pasos anteriores, imprima los Nodes de la nueva Lista Enlazada formada.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// Linked list node
struct Node {
int data;
struct Node* next;
};
// Function to push into 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;
new_node->next = (*head_ref);
// Move the head to point
// to the new node
(*head_ref) = new_node;
}
// Function to print the Linked List
void print(struct Node* head)
{
Node* curr = head;
// Iterate until curr is NULL
while (curr) {
// Print the data
cout << curr->data << " ";
// Move to next
curr = curr->next;
}
}
// Function to create a new Node of
// the Linked List
struct Node* newNode(int x)
{
struct Node* temp
= (struct Node*)malloc(
sizeof(struct Node));
temp->data = x;
temp->next = NULL;
// Return the node created
return temp;
}
// Function used to re-order list
struct Node* reorder(Node* head)
{
// Stores the node of LL
deque<int> v;
Node* curr = head;
// Traverse the LL
while (curr) {
v.push_back(curr->data);
curr = curr->next;
}
// Sort the deque
sort(v.begin(), v.end());
// Node head1 stores the
// head of the new Linked List
Node* head1 = NULL;
Node* prev = NULL;
// Size of new LL
int x = v.size() / 2;
// Loop to make new LL
while (x--) {
int a = v.front();
int b = v.back();
// Difference of squares of
// largest and smallest value
int ans = pow(b, 2) - pow(a, 2);
// Create node with value ans
struct Node* temp = newNode(ans);
if (head1 == NULL) {
head1 = temp;
prev = temp;
}
// Otherwise, update prev
else {
prev->next = temp;
prev = temp;
}
// Pop the front and back node
v.pop_back();
v.pop_front();
}
// Return head of the new LL
return head1;
}
// Driver Code
int main()
{
struct Node* head = NULL;
// Given Linked list
push(&head, 6);
push(&head, 5);
push(&head, 4);
push(&head, 3);
push(&head, 2);
push(&head, 1);
// Function Call
Node* temp = reorder(head);
// Print the new LL formed
print(temp);
return 0;
}
Java
// Java program for the
// above approach
import java.util.*;
class GFG{
// Linked list node
static class Node
{
int data;
Node next;
};
static Node head ;
// Function to push
// into Linked List
static void push(int new_data)
{
// Allocate node
Node new_node = new Node();
// Put in the data
new_node.data = new_data;
new_node.next = head;
// Move the head to point
// to the new node
head = new_node;
}
// Function to print the
// Linked List
static void print(Node head)
{
Node curr = head;
// Iterate until curr
// is null
while (curr != null)
{
// Print the data
System.out.print(curr.data + " ");
// Move to next
curr = curr.next;
}
}
// Function to create a
// new Node of the Linked List
static Node newNode(int x)
{
Node temp = new Node();
temp.data = x;
temp.next = null;
// Return the node
// created
return temp;
}
// Function used to re-order
// list
static Node reorder(Node head)
{
// Stores the node of LL
Deque<Integer> v =
new LinkedList<>();
Node curr = head;
// Traverse the LL
while (curr != null)
{
v.add(curr.data);
curr = curr.next;
}
// Sort the deque
// Collections.sort(v);
// Node head1 stores the
// head of the new Linked
// List
Node head1 = null;
Node prev = null;
// Size of new LL
int x = v.size() / 2;
// Loop to make new LL
while ((x--) > 0)
{
int a = v.peek();
int b = v.getLast();
// Difference of squares of
// largest and smallest value
int ans = (int)(Math.pow(b, 2) -
Math.pow(a, 2));
// Create node with value ans
Node temp = newNode(ans);
if (head1 == null)
{
head1 = temp;
prev = temp;
}
// Otherwise, update prev
else
{
prev.next = temp;
prev = temp;
}
// Pop the front and
// back node
v.removeFirst();
v.removeLast();
}
// Return head of the
// new LL
return head1;
}
// Driver Code
public static void main(String[] args)
{
head = null;
// Given Linked list
push(6);
push(5);
push(4);
push(3);
push(2);
push(1);
// Function Call
Node temp = reorder(head);
// Print the new
// LL formed
print(temp);
}
}
// This code is contributed by Amit Katiyar
Python3
# Python3 program for the # above approach from collections import deque # Linked list node class Node: def __init__(self, x): self.data = x self.next = None # Function to push into Linked List # Function to push into Linked List def push(head_ref, new_data): new_node = Node(new_data) new_node.next = head_ref head_ref = new_node return head_ref # Function to print the Linked List def printt(head): curr = head # Iterate until curr # is None while (curr): # Print the data print(curr.data, end = " ") # Move to next curr = curr.next # Function used to re-order list # Function used to re-order list def reorder(head): # Stores the node of LL arr = [] curr = head while curr: arr.append(curr.data) curr = curr.next arr = sorted(arr) # Sort the deque v = deque() for i in arr: v.append(i) # Node head1 stores the # head of the new Linked List head1 = None prev = None x = len(arr) // 2 while x: a = v.popleft() b = v.pop() # Difference of squares of # largest and smallest value ans = pow(b, 2) - pow(a, 2) temp = Node(ans) if head1 == None: head1 = temp prev = temp else: prev.next = temp prev = temp x -= 1 # Return head of the new LL return head1 # Driver Code if __name__ == '__main__': head = None # Given Linked list head = push(head, 6) head = push(head, 5) head = push(head, 4) head = push(head, 3) head = push(head, 2) head = push(head, 1) # Function Call temp = reorder(head) # Print the new LL formed printt(temp) # This code is contributed by Mohit kumar 29
C#
// C# program for the
// above approach
using System;
using System.Collections.Generic;
class GFG{
// Linked list node
public class Node
{
public int data;
public Node next;
};
static Node head ;
// Function to push
// into Linked List
static void push(int new_data)
{
// Allocate node
Node new_node = new Node();
// Put in the data
new_node.data = new_data;
new_node.next = head;
// Move the head to point
// to the new node
head = new_node;
}
// Function to print the
// Linked List
static void print(Node head)
{
Node curr = head;
// Iterate until curr
// is null
while (curr != null)
{
// Print the data
Console.Write(curr.data + " ");
// Move to next
curr = curr.next;
}
}
// Function to create a
// new Node of the Linked List
static Node newNode(int x)
{
Node temp = new Node();
temp.data = x;
temp.next = null;
// Return the node
// created
return temp;
}
// Function used to re-order
// list
static Node reorder(Node head)
{
// Stores the node of LL
List<int> v =
new List<int>();
Node curr = head;
// Traverse the LL
while (curr != null)
{
v.Add(curr.data);
curr = curr.next;
}
// Sort the deque
// Collections.sort(v);
// Node head1 stores the
// head of the new Linked
// List
Node head1 = null;
Node prev = null;
// Size of new LL
int x = v.Count / 2;
// Loop to make new LL
while ((x--) > 0)
{
int a = v[0];
int b = v[v.Count-1];
// Difference of squares of
// largest and smallest value
int ans = (int)(Math.Pow(b, 2) -
Math.Pow(a, 2));
// Create node with value ans
Node temp = newNode(ans);
if (head1 == null)
{
head1 = temp;
prev = temp;
}
// Otherwise, update prev
else
{
prev.next = temp;
prev = temp;
}
// Pop the front and
// back node
v.RemoveAt(0);
v.RemoveAt(v.Count - 1);
}
// Return head of the
// new LL
return head1;
}
// Driver Code
public static void Main(String[] args)
{
head = null;
// Given Linked list
push(6);
push(5);
push(4);
push(3);
push(2);
push(1);
// Function Call
Node temp = reorder(head);
// Print the new
// LL formed
print(temp);
}
}
// This code is contributed by gauravrajput1
Javascript
<script>
// javascript program for the
// above approach
// Linked list node
class Node {
constructor(val) {
this.data = val;
this.next = null;
}
}
var head;
// Function to push
// into Linked List
function push(new_data) {
// Allocate node
var new_node = new Node();
// Put in the data
new_node.data = new_data;
new_node.next = head;
// Move the head to point
// to the new node
head = new_node;
}
// Function to print the
// Linked List
function print(head) {
var curr = head;
// Iterate until curr
// is null
while (curr != null) {
// Print the data
document.write(curr.data + " ");
// Move to next
curr = curr.next;
}
}
// Function to create a
// new Node of the Linked List
function newNode(x) {
var temp = new Node();
temp.data = x;
temp.next = null;
// Return the node
// created
return temp;
}
// Function used to re-order
// list
function reorder(head) {
// Stores the node of LL
var v = [];
var curr = head;
// Traverse the LL
while (curr != null) {
v.push(curr.data);
curr = curr.next;
}
// Sort the deque
// Collections.sort(v);
// Node head1 stores the
// head of the new Linked
// List
var head1 = null;
var prev = null;
// Size of new LL
var x = v.length / 2;
// Loop to make new LL
while ((x--) > 0) {
var a = v[0];
var b = v[v.length-1];
// Difference of squares of
// largest and smallest value
var ans = parseInt( (Math.pow(b, 2) - Math.pow(a, 2)));
// Create node with value ans
var temp = newNode(ans);
if (head1 == null) {
head1 = temp;
prev = temp;
}
// Otherwise, update prev
else {
prev.next = temp;
prev = temp;
}
// Pop the front and
// back node
v.pop();
v.shift();
}
// Return head of the
// new LL
return head1;
}
// Driver Code
head = null;
// Given Linked list
push(6);
push(5);
push(4);
push(3);
push(2);
push(1);
// Function Call
var temp = reorder(head);
// Print the new
// LL formed
print(temp);
// This code is contributed by todaysgaurav
</script>
Producción:
35 21 7
Complejidad de tiempo: O(N*log N)
Espacio auxiliar: O(N)
Publicación traducida automáticamente
Artículo escrito por kothawade29 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA