算法：编写函数以获取链表中的第N个节点

2021年4月11日15:40:06 发表评论 820 次浏览

本文概述

编写一个GetNth()函数, 该函数接受一个链表和一个整数索引, 并返回存储在该索引位置的节点中的数据值。

例子：

Input:  1->10->30->14, index = 2
Output: 30  
The node at index 2 is 30

算法：

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.

实现

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
     //below 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

C

//C program to find n'th
//node in linked list
#include <assert.h>
#include <stdio.h>
#include <stdlib.h>
  
//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
int GetNth( struct Node* head, int index)
{
  
     struct 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
     struct Node* head = NULL;
  
     //Use push() to construct
     //below 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 ();
}

Java

//Java program to find n'th node in linked list
  
class Node {
     int data;
     Node next;
     Node( int d)
     {
         data = d;
         next = null ;
     }
}
  
class LinkedList {
     Node head; //the head of list
  
     /* Takes index as argument and return data at index*/
     public int GetNth( int index)
     {
         Node current = head;
         int count = 0 ; /* index of Node we are
                           currently looking at */
         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 ( false );
         return 0 ;
     }
  
     /* Given a reference to the head of a list and an int, inserts a new Node on the front of the list. */
     public void push( int new_data)
     {
  
         /* 1. alloc the Node and put data*/
         Node new_Node = new Node(new_data);
  
         /* 2. Make next of new Node as head */
         new_Node.next = head;
  
         /* 3. Move the head to point to new Node */
         head = new_Node;
     }
  
     /* Driver code*/
     public static void main(String[] args)
     {
         /* Start with empty list */
         LinkedList llist = new LinkedList();
  
         /* Use push() to construct below list
            1->12->1->4->1  */
         llist.push( 1 );
         llist.push( 4 );
         llist.push( 1 );
         llist.push( 12 );
         llist.push( 1 );
  
         /* Check the count function */
         System.out.println( "Element at index 3 is "
                            + llist.GetNth( 3 ));
     }
}

python

# A complete working Python program to find n'th node
# in a linked list
  
# Node class
  
  
class Node:
     # Function to initialise the node object
     def __init__( self , data):
         self .data = data  # Assign data
         self . next = None  # Initialize next as null
  
  
# Linked List class contains a Node object
class LinkedList:
  
     # Function to initialize head
     def __init__( self ):
         self .head = None
  
     # This function is in LinkedList class. It inserts
     # a new node at the beginning of Linked List.
  
     def push( self , new_data):
  
         # 1 & 2: Allocate the Node &
         #     Put in the data
         new_node = Node(new_data)
  
         # 3. Make next of new Node as head
         new_node. next = self .head
  
         # 4. Move the head to point to new Node
         self .head = new_node
  
     # Returns data at given index in linked list
     def getNth( self , index):
         current = self .head  # Initialise temp
         count = 0  # Index of current node
  
         # Loop while end of linked list is not reached
         while (current):
             if (count = = index):
                 return current.data
             count + = 1
             current = current. next
  
         # if we get to this line, the caller was asking
         # for a non-existent element so we assert fail
         assert (false)
         return 0
  
  
# Driver Code
if __name__ = = '__main__' :
  
     llist = LinkedList()
  
     # Use push() to construct below list
     # 1->12->1->4->1
     llist.push( 1 )
     llist.push( 4 )
     llist.push( 1 )
     llist.push( 12 )
     llist.push( 1 )
  
     n = 3
     print ( "Element at index 3 is :" , llist.getNth(n))

C#

//C# program to find n'th node in linked list
using System;
using System.Diagnostics;
  
public class Node {
     public int data;
     public Node next;
     public Node( int d)
     {
         data = d;
         next = null ;
     }
}
  
class LinkedList {
     Node head; //the head of list
  
     /* Takes index as argument and return data at index*/
     public int GetNth( int index)
     {
         Node current = head;
         int count = 0; /* index of Node we are
                         currently looking at */
         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 */
         Debug.Assert( false );
         return 0;
     }
  
     /* Given a reference to the head of a list and an int, inserts a new Node on the front of the list. */
     public void push( int new_data)
     {
  
         /* 1. alloc the Node and put data*/
         Node new_Node = new Node(new_data);
  
         /* 2. Make next of new Node as head */
         new_Node.next = head;
  
         /* 3. Move the head to point to new Node */
         head = new_Node;
     }
  
     /* Driver code*/
     public static void Main(String[] args)
     {
         /* Start with empty list */
         LinkedList llist = new LinkedList();
  
         /* Use push() to construct below list
         1->12->1->4->1 */
         llist.push(1);
         llist.push(4);
         llist.push(1);
         llist.push(12);
         llist.push(1);
  
         /* Check the count function */
         Console.WriteLine( "Element at index 3 is "
                           + llist.GetNth(3));
     }
}
  
//This code is contributed by Arnab Kundu

输出如下

Element at index 3 is 4

时间复杂度：上)

方法2-递归

该方法是由

MY_DOOM

.

算法：

Algorithm
getnth(node, n)
1. Initialize count = 0
2. if count==n
     return node->data
3. else
    return getnth(node->next, n-1)

实现

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*/
int GetNth( struct Node* head, int n)
{
     int count = 0;
  
     //if count equal to n return node->data
     if (count == n)
         return head->data;
  
     //recursively decrease n and increase
     //head to next pointer
     return GetNth(head->next, n - 1);
}
  
/* Driver code*/
int main()
{
     /* Start with the empty list */
     struct Node* head = NULL;
  
     /* Use push() to construct below 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 ();
}

Java

//Java program to find n'th node in linked list
//using recursion
class GFG {
  
     /* Link list node */
     static class Node {
         int data;
         Node next;
         Node( int data) { this .data = data; }
     }
  
     /* 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. */
     static Node push(Node head, int new_data)
     {
         /* allocate node */
         Node new_node = new Node(new_data);
  
         /* put in the data */
         new_node.data = new_data;
  
         new_node.next = head;
  
         head = new_node;
  
         return head;
     }
  
     /* Takes head pointer of the linked list and index
         as arguments and return data at index*/
     static int GetNth(Node head, int n)
     {
         int count = 0 ;
         if (head == null ) //edge case - if head is null
             return - 1 ;
         //if count equal too n return node.data
         if (count == n)
             return head.data;
  
         //recursively decrease n and increase
         //head to next pointer
         return GetNth(head.next, n - 1 );
     }
  
     /* Driver code*/
     public static void main(String args[])
     {
         /* Start with the empty list */
         Node head = null ;
  
         /* Use push() to con below list
         1.12.1.4.1 */
         head = push(head, 1 );
         head = push(head, 4 );
         head = push(head, 1 );
         head = push(head, 12 );
         head = push(head, 1 );
  
         /* Check the count function */
         System.out.printf( "Element at index 3 is %d" , GetNth(head, 3 ));
     }
}
//This code is contributed by Arnab Kundu

Python3

# Python3 program to find n'th node in
# linked list using recursion
  
  
class Node:
     def __init__( self , data):
         self .data = data
         self . next = None
  
  
class LinkedList:
     def __init__( self ):
         self .head = None
  
     ''' 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. '''
  
     def push( self , new_data):  # make new node and add
                               # into LinkedList
         new_node = Node(new_data)
         new_node. next = self .head
         self .head = new_node
  
     def getNth( self , llist, position):
  
         # call recursive method
         llist.getNthNode( self .head, position, llist)
  
     # recursive method to find Nth Node
     def getNthNode( self , head, position, llist):
         count = 0  # initialize count
         if (head):
             if count = = position:  # if count is equal to position, # it means we have found the position
                 print (head.data)
             else :
                 llist.getNthNode(head. next , position - 1 , llist)
         else :  # if head doesn't exist we have
               # traversed the LinkedList
             print ( 'Index Doesn\'t exist' )
  
  
# Driver Code
if __name__ = = "__main__" :
     llist = LinkedList()
     llist.push( 1 )
     llist.push( 4 )
     llist.push( 1 )
     llist.push( 12 )
     llist.push( 1 )
     # llist.getNth(llist, int(input()))
     # Enter the node position here
     # first argument is instance of LinkedList
  
     print ( "Element at Index 3 is" , end = " " )
     llist.getNth(llist, 3 )
  
# This code is contributed by Yogesh Joshi

C#

//C# program to find n'th node in
//linked list using recursion
using System;
  
class GFG {
  
     /* Link list node */
     public class Node {
         public int data;
         public Node next;
         public Node( int data) { this .data = data; }
     }
  
     /* 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. */
     static Node push(Node head, int new_data)
     {
         /* allocate node */
         Node new_node = new Node(new_data);
  
         /* put in the data */
         new_node.data = new_data;
  
         new_node.next = head;
  
         head = new_node;
  
         return head;
     }
  
     /* Takes head pointer of the linked list and index
         as arguments and return data at index*/
     static int GetNth(Node head, int n)
     {
         //Base Condition
         if (head == null )
             return -1;
  
         int count = 0;
  
         //if count equal too n return node.data
         //Test Condition
         if (count == n)
             return head.data;
  
         //recursively decrease n and increase
         //head to next pointer
         return GetNth(head.next, n - 1);
     }
  
     /* Driver code*/
     public static void Main()
     {
         /* Start with the empty list */
         Node head = null ;
  
         /* Use push() to con below list
         1.12.1.4.1 */
         head = push(head, 1);
         head = push(head, 4);
         head = push(head, 1);
         head = push(head, 12);
         head = push(head, 1);
  
         /* Check the count function */
         Console.Write( "Element at index 3 is {0}" , GetNth(head, 3));
     }
}
/*Code improvement by Aishwarya Mittal*/
/* This code contributed by PrinciRaj1992 */

输出如下

Element at index 3 is 4

时间复杂度：O(n)

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