Dada una lista enlazada individualmente, su tarea es eliminar cada K-ésimo Node de la lista enlazada. Suponga que K siempre es menor o igual que la longitud de la lista enlazada.
Ejemplos:
Input: 1->2->3->4->5->6->7->8 k = 3 Output: 1->2->4->5->7->8 As 3 is the k-th node after its deletion list would be 1->2->4->5->6->7->8 And now 4 is the starting node then from it, 6 would be the k-th node. So no other kth node could be there.So, final list is: 1->2->4->5->7->8. Input: 1->2->3->4->5->6 k = 1 Output: Empty list All nodes need to be deleted
La idea es recorrer la lista desde el principio y realizar un seguimiento de los Nodes visitados después de la última eliminación. Cada vez que el conteo se convierte en k, elimine el Node actual y restablezca el conteo a 0.
Traverse list and do following (a) Count node before deletion. (b) If (count == k) that means current node is to be deleted. (i) Delete current node i.e. do // assign address of next node of // current node to the previous node // of the current node. prev->next = ptr->next i.e. (ii) Reset count as 0, i.e., do count = 0. (c) Update prev node if count != 0 and if count is 0 that means that node is a starting point. (d) Update ptr and continue until all k-th node gets deleted.
A continuación se muestra la implementación.
Java
// Java program to delete every k-th // Node of a singly linked list. class GFG{ // Linked list Node static class Node { int data; Node next; } // To remove complete list (Needed // for case when k is 1) static Node freeList(Node node) { while (node != null) { Node next = node.next; node = next; } return node; } // Deletes every k-th node and // returns head of modified list. static Node deleteKthNode(Node head, int k) { // If linked list is empty if (head == null) return null; if (k == 1) { head = freeList(head); return null; } // Initialize ptr and prev before // starting traversal. Node ptr = head, prev = null; // Traverse list and delete // every k-th node int count = 0; while (ptr != null) { // Increment Node count count++; // Check if count is equal to k // if yes, then delete current Node if (k == count) { // Put the next of current Node in // the next of previous Node prev.next = ptr.next; // Set count = 0 to reach further // k-th Node count = 0; } // Update prev if count is not 0 if (count != 0) prev = ptr; ptr = prev.next; } return head; } // Function to print linked list static void displayList(Node head) { Node temp = head; while (temp != null) { System.out.print(temp.data + " "); temp = temp.next; } } // Utility function to create a // new node. static Node newNode(int x) { Node temp = new Node(); temp.data = x; temp.next = null; return temp; } // Driver Code public static void main(String args[]) { // Start with the empty list Node head = newNode(1); head.next = newNode(2); head.next.next = newNode(3); head.next.next.next = newNode(4); head.next.next.next.next = newNode(5); head.next.next.next.next.next = newNode(6); head.next.next.next.next.next.next = newNode(7); head.next.next.next.next.next.next.next = newNode(8); int k = 3; head = deleteKthNode(head, k); displayList(head); } } // This code is contributed by Arnab Kundu
Producción:
1 2 4 5 7 8
Complejidad de tiempo: O(n)
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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