在C和Java中找到给定链表的中间部分

2021年3月12日13:02:29 发表评论 806 次浏览

本文概述

给定一个单链表, 找到链表的中间。例如, 如果给定的链表为1-> 2-> 3-> 4-> 5, 则输出应为3。

如果有偶数节点, 那么将有两个中间节点, 我们需要打印第二个中间元素。例如, 如果给定的链表为1-> 2-> 3-> 4-> 5-> 6, 则输出应为4。

推荐：请在"

实践

首先, 在继续解决方案之前。

方法1：

遍历整个链表并计算编号。节点。现在再次遍历列表, 直到count / 2并返回count / 2处的节点。

方法2：

使用两个指针遍历链表。将一个指针移动一个, 另一个指针移动两个。当快速指针到达末尾时, 慢速指针将到达链表的中间。

下图显示了printMiddle函数如何在代码中工作：

在C和Java中找到给定链表的中间部分1

C

#include<stdio.h> 
#include<stdlib.h> 
  
/* Link list node */
struct Node 
{ 
     int data; 
     struct Node* next; 
}; 
  
/* Function to get the middle of the linked list*/
void printMiddle( struct Node *head) 
{ 
     struct Node *slow_ptr = head; 
     struct Node *fast_ptr = head; 
  
     if (head!=NULL) 
     { 
         while (fast_ptr != NULL && fast_ptr->next != NULL) 
         { 
             fast_ptr = fast_ptr->next->next; 
             slow_ptr = slow_ptr->next; 
         } 
         printf ( "The middle element is [%d]\n\n" , slow_ptr->data); 
     } 
} 
  
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; 
} 
  
// A utility function to print a given linked list 
void printList( struct Node *ptr) 
{ 
     while (ptr != NULL) 
     { 
         printf ( "%d->" , ptr->data); 
         ptr = ptr->next; 
     } 
     printf ( "NULL\n" ); 
} 
  
/* Driver program to test above function*/
int main() 
{ 
     /* Start with the empty list */
     struct Node* head = NULL; 
     int i; 
  
     for (i=5; i>0; i--) 
     { 
         push(&head, i); 
         printList(head); 
         printMiddle(head); 
     } 
  
     return 0; 
}

C ++

#include<bits/stdc++.h> 
using namespace std; 
  
// Struct 
struct Node 
{ 
     int data; 
     struct Node* next; 
}; 
  
/* Function to get the middle of the linked list*/
void printMiddle( struct Node *head) 
{ 
     struct Node *slow_ptr = head; 
     struct Node *fast_ptr = head; 
  
     if (head!=NULL) 
     { 
         while (fast_ptr != NULL && fast_ptr->next != NULL) 
         { 
             fast_ptr = fast_ptr->next->next; 
             slow_ptr = slow_ptr->next; 
         } 
         printf ( "The middle element is [%d]\n\n" , slow_ptr->data); 
     } 
} 
  
// Function to add a new node 
void push( struct Node** head_ref, int new_data) 
{ 
     /* allocate node */
     struct 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; 
} 
  
// A utility function to print a given linked list 
void printList( struct Node *ptr) 
{ 
     while (ptr != NULL) 
     { 
         printf ( "%d->" , ptr->data); 
         ptr = ptr->next; 
     } 
     printf ( "NULL\n" ); 
} 
  
// Driver Code
int main() 
{ 
     // Start with the empty list 
     struct Node* head = NULL; 
      
     // Iterate and add element 
     for ( int i=5; i>0; i--) 
     { 
         push(&head, i); 
         printList(head); 
         printMiddle(head); 
     } 
  
     return 0; 
}

Java

// Java program to find middle of linked list
class LinkedList
{
     Node head; // head of linked list
  
     /* Linked list node */
     class Node
     {
         int data;
         Node next;
         Node( int d)
         {
             data = d;
             next = null ;
         }
     }
  
     /* Function to print middle of linked list */
     void printMiddle()
     {
         Node slow_ptr = head;
         Node fast_ptr = head;
         if (head != null )
         {
             while (fast_ptr != null && fast_ptr.next != null )
             {
                 fast_ptr = fast_ptr.next.next;
                 slow_ptr = slow_ptr.next;
             }
             System.out.println( "The middle element is [" +
                                 slow_ptr.data + "] \n" );
         }
     }
  
     /* Inserts a new Node at front of the list. */
     public void push( int new_data)
     {
         /* 1 & 2: Allocate the Node &
                   Put in the data*/
         Node new_node = new Node(new_data);
  
         /* 3. Make next of new Node as head */
         new_node.next = head;
  
         /* 4. Move the head to point to new Node */
         head = new_node;
     }
  
     /* This function prints contents of linked list
        starting from  the given node */
     public void printList()
     {
         Node tnode = head;
         while (tnode != null )
         {
             System.out.print(tnode.data+ "->" );
             tnode = tnode.next;
         }
         System.out.println( "NULL" );
     }
  
     public static void main(String [] args)
     {
         LinkedList llist = new LinkedList();
         for ( int i= 5 ; i> 0 ; --i)
         {
             llist.push(i);
             llist.printList();
             llist.printMiddle();
         }
     }
}
// This code is contributed by Rajat Mishra

输出如下： 

5->NULL
The middle element is [5]

4->5->NULL
The middle element is [5]

3->4->5->NULL
The middle element is [4]

2->3->4->5->NULL
The middle element is [4]

1->2->3->4->5->NULL
The middle element is [3]

方法3：

将中部元素初始化为head并将计数器初始化为0。从头开始遍历列表, 同时遍历递增计数器, 并在计数器为奇数时从中到中>下一个。因此, 中间部分只会移动列表总长度的一半。

感谢Narendra Kangralkar提出了这种方法。

C

#include <stdio.h>
#include <stdlib.h>
  
/* Link list node */
struct node {
     int data;
     struct node* next;
};
  
/* Function to get the middle of the linked list*/
void printMiddle( struct node* head)
{
     int count = 0;
     struct node* mid = head;
  
     while (head != NULL) {
         /* update mid, when 'count' is odd number */
         if (count & 1)
             mid = mid->next;
  
         ++count;
         head = head->next;
     }
  
     /* if empty list is provided */
     if (mid != NULL)
         printf ( "The middle element is [%d]\n\n" , mid->data);
}
  
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;
}
  
// A utility function to print a given linked list
void printList( struct node* ptr)
{
     while (ptr != NULL) {
         printf ( "%d->" , ptr->data);
         ptr = ptr->next;
     }
     printf ( "NULL\n" );
}
  
/* Driver program to test above function*/
int main()
{
     /* Start with the empty list */
     struct node* head = NULL;
     int i;
  
     for (i = 5; i > 0; i--) {
         push(&head, i);
         printList(head);
         printMiddle(head);
     }
  
     return 0;
}

Java

class GFG{
  
static Node head;
  
// Link list node
class Node
{
     int data;
     Node next;
  
     // Constructor
     public Node(Node next, int data)
     {
         this .data = data;
         this .next = next;
     }
}
  
// Function to get the middle of 
// the linked list
void printMiddle(Node head)
{
     int count = 0 ;
     Node mid = head;
  
     while (head != null )
     {
  
         // Update mid, when 'count' 
         // is odd number 
         if ((count % 2 ) == 1 )
             mid = mid.next;
  
         ++count;
         head = head.next;
     }
  
     // If empty list is provided 
     if (mid != null )
         System.out.println( "The middle element is [" +
                             mid.data + "]\n" );
}
  
void push(Node head_ref, int new_data)
{
      
     // Allocate node 
     Node new_node = new Node(head_ref, new_data);
  
     // Move the head to point to the new node 
     head = new_node;
}
  
// A utility function to print a 
// given linked list
void printList(Node head)
{
     while (head != null )
     {
         System.out.print(head.data + "-> " );
         head = head.next;
     }
     System.out.println( "null" );
}
  
// Driver code
public static void main(String[] args)
{
     GFG ll = new GFG();
  
     for ( int i = 5 ; i > 0 ; i--) 
     {
         ll.push(head, i);
         ll.printList(head);
         ll.printMiddle(head);
     }
}
}
  
// This code is contributed by mark_3

输出如下

5->NULL
The middle element is [5]

4->5->NULL
The middle element is [5]

3->4->5->NULL
The middle element is [4]

2->3->4->5->NULL
The middle element is [4]

1->2->3->4->5->NULL
The middle element is [3]

如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。

木子山

发表评论

:?: :razz: :sad: :evil: :!: :smile: :oops: :grin: :eek: :shock: :???: :cool: :lol: :mad: :twisted: :roll: :wink: :idea: :arrow: :neutral: :cry: :mrgreen: