在链表中找到分数(或 n/k – th)节点
给定一个单向链表和一个数字 k,编写一个函数来查找第 (n/k) 个元素,其中 n 是列表中元素的数量。在小数的情况下,我们需要考虑 ceil 值。
例子:
Input : list = 1->2->3->4->5->6
k = 2
Output : 3
Since n = 6 and k = 2, we print (6/2)-th node
which is 3.
Input : list = 2->7->9->3->5
k = 3
Output : 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fractionalNode 并分别用 null 和 head 初始化它们。
- 对于临时指针的每 k 次跳转,使小数节点指针跳转一次。
C++
// C++ program to find fractional node in a linked list
#include
/* 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("\n");
}
/* 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("\nFractional node is ");
printf("%d\n", 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 prList(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 = ' ')
prList(head)
answer = fractionalNodes(head, k)
print("\nFractional 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
输出:
List is 1 2 3 4 5
Fractional node is 3
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