Escriba una función GetNth() que tome una lista enlazada y un índice entero y devuelva el valor de datos almacenado en el Node en esa posición de índice.
Ejemplo:
Input: 1->10->30->14, index = 2 Output: 30 The node at index 2 is 30
Algoritmo:
1. Initialize count = 0 2. Loop through the link list a. If count is equal to the passed index then return current node b. Increment count c. change current to point to next of the current.
Implementación:
C++
// C++ program to find n'th // node in linked list #include <assert.h> #include <bits/stdc++.h> using namespace std; // Link list node class Node { public: int data; Node* next; }; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the list. */ void 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; } // Takes head pointer of // the linked list and index // as arguments and return // data at index int GetNth(Node* head, int index) { Node* current = head; // The index of the // node we're currently // looking at int count = 0; while (current != NULL) { if (count == index) return (current->data); count++; current = current->next; } /* If we get to this line, the caller was asking for a non-existent element so we assert fail */ assert(0); } // Driver Code int main() { // Start with the empty list Node* head = NULL; // Use push() to construct list // 1->12->1->4->1 push(&head, 1); push(&head, 4); push(&head, 1); push(&head, 12); push(&head, 1); // Check the count function cout << "Element at index 3 is " << GetNth(head, 3); return 0; } // This code is contributed by rathbhupendra
Producción:
Element at index 3 is 4
Complejidad de tiempo: O(n)
Complejidad espacial: O(1) usando solo variables constantes
Método 2: con recursividad:
este método lo aporta MY_DOOM .
Algoritmo:
getnth(node,n) 1. Initialize count = 0 2. if count==n return node->data 3. else return getnth(node->next,n-1)
Implementación:
C++
// C++ program to find n'th node in // linked list using recursion #include <bits/stdc++.h> using namespace std; // Link list node struct Node { int data; struct Node* next; }; /* Given a reference (pointer to pointer) to the head of a list and an int, push a new node on the front of the 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; } /* Takes head pointer of the linked list and index as arguments and return data at index. (Don't use another variable)*/ int GetNth(struct Node* head, int n) { // If length of the list is less // than the given index, return -1 if (head == NULL) return -1; // If n equal to 0 return node->data if (n == 0) return head->data; // Increase head to next pointer // n - 1: decrease the number of // recursions until n = 0 return GetNth(head->next, n - 1); } // Driver code int main() { // Start with the empty list struct Node* head = NULL; // Use push() to construct list // 1->12->1->4->1 push(&head, 1); push(&head, 4); push(&head, 1); push(&head, 12); push(&head, 1); // Check the count function printf("Element at index 3 is %d", GetNth(head, 3)); getchar(); }
Producción:
Element at index 3 is 4
Complejidad de tiempo: O(n)
Complejidad espacial : O (n) ya que usa espacio constante para crear Nodes
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Publicación traducida automáticamente
Artículo escrito por GeeksforGeeks-1 y traducido por Barcelona Geeks. The original can be accessed here. Licence: CCBY-SA