给定一个单链表和一个数字k，编写一个函数来查找第（n/k）个元素，其中n是列表中的元素数。在小数的情况下，我们需要考虑CEIL值。 例如：
null
Input : list = 1->2->3->4->5->6 k = 2Output : 3Since n = 6 and k = 2, we print (6/2)-th node which is 3.Input : list = 2->7->9->3->5 k = 3Output : 7 Since n is 5 and k is 3, we print ceil(5/3)-th node which is 2nd node, i.e., 7.
- 取两个指针temp和分馏节点，分别用null和head初始化它们。
- 对于临时指针的每k次跳转，对分馏节点指针进行一次跳转。
C++
// C++ program to find fractional node in a linked list #include <bits/stdc++.h> /* Linked list node */ struct Node { int data; Node* next; }; /* Function to create a new node with given data */ Node* newNode( int data) { Node* new_node = new Node; new_node->data = data; new_node->next = NULL; return new_node; } /* Function to find fractional node in the linked list */ Node* fractionalNodes(Node* head, int k) { // Corner cases if (k <= 0 || head == NULL) return NULL; Node* fractionalNode = NULL; // Traverse the given list int i = 0; for (Node* temp = head; temp != NULL; temp = temp->next) { // For every k nodes, we move fractionalNode one // step ahead. if (i % k == 0) { // First time we see a multiple of k if (fractionalNode == NULL) fractionalNode = head; else fractionalNode = fractionalNode->next; } i++; } return fractionalNode; } // A utility function to print a linked list void printList(Node* node) { while (node != NULL) { printf ( "%d " , node->data); node = node->next; } printf ( "" ); } /* Driver program to test above function */ int main( void ) { 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); int k = 2; printf ( "List is " ); printList(head); Node* answer = fractionalNodes(head, k); printf ( "Fractional node is " ); printf ( "%d" , answer->data); return 0; } |
JAVA
// Java program to find fractional node in // a linked list public class FractionalNodell { /* Linked list node */ static class Node{ int data; Node next; //Constructor Node ( int data){ this .data = data; } } /* Function to find fractional node in the linked list */ static Node fractionalNodes(Node head, int k) { // Corner cases if (k <= 0 || head == null ) return null ; Node fractionalNode = null ; // Traverse the given list int i = 0 ; for (Node temp = head; temp != null ; temp = temp.next){ // For every k nodes, we move // fractionalNode one step ahead. if (i % k == 0 ){ // First time we see a multiple of k if (fractionalNode == null ) fractionalNode = head; else fractionalNode = fractionalNode.next; } i++; } return fractionalNode; } // A utility function to print a linked list static void printList(Node node) { while (node != null ) { System.out.print(node.data+ " " ); node = node.next; } System.out.println(); } /* Driver program to test above function */ public static void main(String[] args) { Node head = new Node( 1 ); head.next = new Node( 2 ); head.next.next = new Node( 3 ); head.next.next.next = new Node( 4 ); head.next.next.next.next = new Node( 5 ); int k = 2 ; System.out.print( "List is " ); printList(head); Node answer = fractionalNodes(head, k); System.out.println( "Fractional node is " + answer.data); } } // This code is contributed by Sumit Ghosh |
Python3
# Python3 program to find fractional node # in a linked list import math # Linked list node class Node: def __init__( self , data): self .data = data self . next = None # Function to create a new node # with given data def newNode(data): new_node = Node(data) new_node.data = data new_node. next = None return new_node # Function to find fractional node # in the linked list def fractionalNodes(head, k): # Corner cases if (k < = 0 or head = = None ): return None fractionalNode = None # Traverse the given list i = 0 temp = head while (temp ! = None ): # For every k nodes, we move # fractionalNode one step ahead. if (i % k = = 0 ): # First time we see a multiple of k if (fractionalNode = = None ): fractionalNode = head else : fractionalNode = fractionalNode. next i = i + 1 temp = temp. next return fractionalNode # A utility function to print a linked list def printList(node): while (node ! = None ): print (node.data, end = ' ' ) node = node. next # Driver Code if __name__ = = '__main__' : 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 ) k = 2 print ( "List is" , end = ' ' ) printList(head) answer = fractionalNodes(head, k) print ( "Fractional node is" , end = ' ' ) print (answer.data) # This code is contributed by Srathore |
C#
// C# program to find fractional node in // a linked list using System; public class FractionalNodell { /* Linked list node */ public class Node { public int data; public Node next; //Constructor public Node ( int data) { this .data = data; } } /* Function to find fractional node in the linked list */ static Node fractionalNodes(Node head, int k) { // Corner cases if (k <= 0 || head == null ) return null ; Node fractionalNode = null ; // Traverse the given list int i = 0; for (Node temp = head; temp != null ; temp = temp.next) { // For every k nodes, we move // fractionalNode one step ahead. if (i % k == 0) { // First time we see a multiple of k if (fractionalNode == null ) fractionalNode = head; else fractionalNode = fractionalNode.next; } i++; } return fractionalNode; } // A utility function to print a linked list static void printList(Node node) { while (node != null ) { Console.Write(node.data+ " " ); node = node.next; } Console.WriteLine(); } /* Driver code */ public static void Main(String[] args) { Node head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); int k =2; Console.Write( "List is " ); printList(head); Node answer = fractionalNodes(head, k); Console.WriteLine( "Fractional node is " + answer.data); } } // This code is contributed by Rajput-Ji |
Javascript
<script> // JavaScript program to find fractional node in // a linked list /* Linked list node */ class Node { constructor(val) { this .data = val; this .next = null ; } } /* * Function to find fractional node in the linked list */ function fractionalNodes(head , k) { // Corner cases if (k <= 0 || head == null ) return null ; var fractionalNode = null ; // Traverse the given list var i = 0; for (temp = head; temp != null ; temp = temp.next) { // For every k nodes, we move // fractionalNode one step ahead. if (i % k == 0) { // First time we see a multiple of k if (fractionalNode == null ) fractionalNode = head; else fractionalNode = fractionalNode.next; } i++; } return fractionalNode; } // A utility function to print a linked list function printList(node) { while (node != null ) { document.write(node.data + " " ); node = node.next; } document.write(); } /* Driver program to test above function */ var head = new Node(1); head.next = new Node(2); head.next.next = new Node(3); head.next.next.next = new Node(4); head.next.next.next.next = new Node(5); var k = 2; document.write( "List is " ); printList(head); var answer = fractionalNodes(head, k); document.write( "<br/>Fractional node is " + answer.data); // This code contributed by Rajput-Ji </script> |
输出：
List is 1 2 3 4 5 Fractional node is 3
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