本文概述

• C
• C ++
• Java
• C
• Java

【在C和Java中找到给定链表的中间部分】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：

文章图片
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 = https://www.lsbin.com/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 = https://www.lsbin.com/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 = https://www.lsbin.com/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 fromthe 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-> NULLThe middle element is [5]4-> 5-> NULLThe middle element is [5]3-> 4-> 5-> NULLThe middle element is [4]2-> 3-> 4-> 5-> NULLThe middle element is [4]1-> 2-> 3-> 4-> 5-> NULLThe 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 = https://www.lsbin.com/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 = https://www.lsbin.com/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-> NULLThe middle element is [5]4-> 5-> NULLThe middle element is [5]3-> 4-> 5-> NULLThe middle element is [4]2-> 3-> 4-> 5-> NULLThe middle element is [4]1-> 2-> 3-> 4-> 5-> NULLThe 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。

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