用于查找给定链表的中间元素的 C++ 程序

给定一个单链表，找到链表的中间。例如，如果给定的链表是 1->2->3->4->5，那么输出应该是 3。
如果有偶数节点，那么就会有两个中间节点，我们需要打印第二个中间元素。例如，如果给定的链表是 1->2->3->4->5->6，那么输出应该是 4。

方法一：
遍历整个链表并计算编号。的节点。现在再次遍历列表直到 count/2 并返回 count/2 处的节点。

方法二：
使用两个指针遍历链表。将一个指针移动一格，将其他指针移动二格。当快指针到达末尾时，慢指针将到达链表的中间。

下图显示了 printMiddle函数在代码中的工作方式：

C++

// C++ program for the above approach
  
#include 
using namespace std;
  
class Node{
    public:
        int data;
        Node *next;
};
  
class NodeOperation{
  public: 
    
    // Function to add a new node
    void pushNode(class Node** head_ref,int data_val){
        
        // Allocate node 
        class Node *new_node = new Node();
          
         // Put in the data 
        new_node->data = data_val;
          
        // 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 printNode(class Node *head){
        while(head != NULL){
            cout <data << "->";
            head = head->next;
        }
        cout << "NULL" << endl;
    }
      
    void printMiddle(class 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;
            }
            cout << "The middle element is [" << slow_ptr->data << "]" << endl;
        }
    }
};
  
// Driver Code
int main(){
    class Node *head = NULL;
    class NodeOperation *temp = new NodeOperation();
    for(int i=5; i>0; i--){
        temp->pushNode(&head, i);
        temp->printNode(head);
        temp->printMiddle(head);
    }
    return 0;
}

输出

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]

方法三：
将 mid 元素初始化为 head 并将计数器初始化为 0。从 head 遍历列表，同时遍历递增计数器并在计数器为奇数时将 mid 更改为 mid->next。所以中间只会移动列表总长度的一半。
感谢 Narendra Kangralkar 提出这种方法。

C++

#include 
using namespace std;
  
// 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]
  
", 
                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
");
}
  
// Driver code
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;
}
  
// This code is contributed by ac121102

输出

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]

有关详细信息，请参阅有关查找给定链接列表的中间的完整文章！