Dada una lista enlazada, la tarea es encontrar la suma de todos los Nodes que son mayores que el Node contiguo. Tenga en cuenta que para el último Node de la lista enlazada que no tiene ningún Node al lado, debe ser mayor que el primer Node para que contribuya a la suma.
Ejemplos:
Entrada: 9 -> 2 -> 3 -> 5 -> 4 -> 6 -> 8
Salida: 14
9 + 5 = 14
Entrada: 2 -> 1 -> 5 -> 7
Salida: 9
2 + 7 = 9
Enfoque: recorra toda la lista enlazada y para cada Node, si el Node es mayor que el siguiente, agréguelo a la suma. Para el último Node, compárelo con el encabezado de la lista enlazada, si el último Node es mayor que el encabezado, agréguelo a la suma. Imprime la suma al final.
A continuación se muestra la implementación del enfoque anterior:
C++
// C++ implementation of the approach #include <iostream> using namespace std; // Represents node of the linked list struct Node { int data; Node* next; }; // Function to insert a node at the // end of the linked list void insert(Node** root, int item) { Node *ptr = *root, *temp = new Node; temp->data = item; temp->next = NULL; if (*root == NULL) *root = temp; else { while (ptr->next != NULL) ptr = ptr->next; ptr->next = temp; } } // Function to return the sum of the nodes // which are greater than the node next to them int sum(Node* root) { // If there are no nodes if (root == NULL) return 0; int sm = 0; Node* ptr = root; while (ptr->next != NULL) { // If the node is greater than the next node if (ptr->data > ptr->next->data) sm += ptr->data; ptr = ptr->next; } // For the last node if (ptr->data > root->data) sm += ptr->data; // Return the sum return sm; } // Driver code int main() { Node* root = NULL; insert(&root, 9); insert(&root, 2); insert(&root, 3); insert(&root, 5); insert(&root, 4); insert(&root, 6); insert(&root, 8); cout << sum(root) << endl; return 0; }
Java
// Java implementation of the approach class GFG { // Represents node of the linked list static class Node { int data; Node next; }; // Function to insert a node at the // end of the linked list static Node insert(Node root, int item) { Node ptr = root, temp = new Node(); temp.data = item; temp.next = null; if (root == null) root = temp; else { while (ptr.next != null) ptr = ptr.next; ptr.next = temp; } return root; } // Function to return the sum of the nodes // which are greater than the node next to them static int sum(Node root) { // If there are no nodes if (root == null) return 0; int sm = 0; Node ptr = root; while (ptr.next != null) { // If the node is greater than the next node if (ptr.data > ptr.next.data) sm += ptr.data; ptr = ptr.next; } // For the last node if (ptr.data > root.data) sm += ptr.data; // Return the sum return sm; } // Driver code public static void main(String args[]) { Node root = null; root=insert(root, 9); root=insert(root, 2); root=insert(root, 3); root=insert(root, 5); root=insert(root, 4); root=insert(root, 6); root=insert(root, 8); System.out.print( sum(root) ); } } // This code is contributed by Arnab Kundu
Python3
# Python3 implementation of the approach import math # Represents node of the linked list class Node: def __init__(self, data): self.data = data self.next = None # Function to root=insert a node at the # end of the linked list def insert(root, item): ptr = root temp = Node(item); temp.data = item; temp.next = None; if (root == None): root = temp; else: while (ptr.next != None): ptr = ptr.next; ptr.next = temp; return root # Function to return the sum of the nodes # which are greater than the node next to them def sum(root): # If there are no nodes if (root == None): return 0; sm = 0; ptr = root; while (ptr.next != None): # If the node is greater than the next node if (ptr.data > ptr.next.data): sm += ptr.data; ptr = ptr.next; # For the last node if (ptr.data > root.data): sm += ptr.data; # Return the sum return sm; # Driver code if __name__=='__main__': root = None; root = insert(root, 9); root = insert(root, 2); root = insert(root, 3); root = insert(root, 5); root = insert(root, 4); root = insert(root, 6); root = insert(root, 8); print(sum(root)) # This code is contributed by Srathore
C#
// C# implementation of the approach using System; class GFG { // Represents node of the linked list public class Node { public int data; public Node next; }; // Function to insert a node at the // end of the linked list static Node insert(Node root, int item) { Node ptr = root, temp = new Node(); temp.data = item; temp.next = null; if (root == null) root = temp; else { while (ptr.next != null) ptr = ptr.next; ptr.next = temp; } return root; } // Function to return the sum of the nodes // which are greater than the node next to them static int sum(Node root) { // If there are no nodes if (root == null) return 0; int sm = 0; Node ptr = root; while (ptr.next != null) { // If the node is greater than the next node if (ptr.data > ptr.next.data) sm += ptr.data; ptr = ptr.next; } // For the last node if (ptr.data > root.data) sm += ptr.data; // Return the sum return sm; } // Driver code public static void Main(String []args) { Node root = null; root = insert(root, 9); root = insert(root, 2); root = insert(root, 3); root = insert(root, 5); root = insert(root, 4); root = insert(root, 6); root = insert(root, 8); Console.Write( sum(root) ); } } // This code contributed by Rajput-Ji
Javascript
<script> // Javascript implementation of the approach // Represents node of the linked list class Node { constructor() { this.data = 0; this.next = null; } } // Function to insert a node at the // end of the linked list function insert( root, item) { var ptr = root, temp = new Node(); temp.data = item; temp.next = null; if (root == null) root = temp; else { while (ptr.next != null) ptr = ptr.next; ptr.next = temp; } return root; } // Function to return the sum of the nodes // which are greater than the node next to them function sum( root) { // If there are no nodes if (root == null) return 0; let sm = 0; var ptr = root; while (ptr.next != null) { // If the node is greater than the next node if (ptr.data > ptr.next.data) sm += ptr.data; ptr = ptr.next; } // For the last node if (ptr.data > root.data) sm += ptr.data; // Return the sum return sm; } // Driver Code var root = null; root=insert(root, 9); root=insert(root, 2); root=insert(root, 3); root=insert(root, 5); root=insert(root, 4); root=insert(root, 6); root=insert(root, 8); document.write( sum(root) ); // This code is contributed by jana_sayantan. </script>
14
Complejidad de tiempo : O(n) donde n es el tamaño de la lista enlazada
Publicación traducida automáticamente
Artículo escrito por Premdeep Toppo y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA