用于删除链表的每个第 K 个节点的 C++ 程序

给定一个单链表，你的任务是删除链表的每个第 K 个节点。假设 K 总是小于或等于 Linked List 的长度。
例子 ：

Input: 1->2->3->4->5->6->7->8  
        k = 3
Output: 1->2->4->5->7->8
As 3 is the k-th node after its deletion list 
would be 1->2->4->5->6->7->8
And now 4 is the starting node then from it, 6 
would be the k-th node. So no other kth node 
could be there.So, final list is:
1->2->4->5->7->8.

Input: 1->2->3->4->5->6  
       k = 1
Output: Empty list 
All nodes need to be deleted

这个想法是从头开始遍历列表并跟踪上次删除后访问的节点。每当count变为k时，删除当前节点并将count重置为0。

Traverse list and do following
   (a) Count node before deletion.
   (b) If (count == k) that means current 
        node is to be deleted.
      (i)  Delete current node i.e. do

          //  assign address of next node of 
          // current node to the previous node
          // of the current node.
          prev->next = ptr->next i.e.

       (ii) Reset count as 0, i.e., do count = 0.
   (c) Update prev node if count != 0 and if
       count is 0 that means that node is a
       starting point.
   (d) Update ptr and continue until all 
       k-th node gets deleted.

下面是实现。

C++

// C++ program to delete every k-th 
// Node of a singly linked list.
#include
using namespace std;
  
// Linked list Node 
struct Node
{
    int data;
    struct Node* next;
};
  
// To remove complete list (Needed 
// for case when k is 1)
void freeList(Node *node)
{
    while (node != NULL)
    {
        Node *next = node->next;
        delete (node);
        node  = next;
    }
}
  
// Deletes every k-th node and returns head
// of modified list.
Node *deleteKthNode(struct Node *head, 
                    int k)
{
    // If linked list is empty
    if (head == NULL)
        return NULL;
  
    if (k == 1)
    {
       freeList(head);
       return NULL;
    }
  
    // Initialize ptr and prev before 
    // starting traversal.
    struct Node *ptr = head, *prev = NULL;
  
    // Traverse list and delete every 
    // k-th node
    int count = 0;
    while (ptr != NULL)
    {
        // Increment Node count
        count++;
  
        // Check if count is equal to k
        // if yes, then delete current Node
        if (k == count)
        {
            // Put the next of current Node in
            // the next of previous Node
            delete(prev->next);
            prev->next = ptr->next;
  
            // Set count = 0 to reach further
            // k-th Node
            count = 0;
        }
  
        // Update prev if count is not 0
        if (count != 0)
            prev = ptr;
  
        ptr = prev->next;
    }
  
    return head;
}
  
// Function to print linked list
void displayList(struct Node *head)
{
    struct Node *temp = head;
    while (temp != NULL)
    {
        cout << temp->data << " ";
        temp = temp->next;
    }
}
  
// Utility function to create 
// a new node.
struct Node *newNode(int x)
{
    Node *temp = new Node;
    temp->data = x;
    temp->next = NULL;
    return temp;
}
  
// Driver code
int main()
{
    // Start with the empty list 
    struct 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);
    head->next->next->next->next->next = 
    newNode(6);
    head->next->next->next->next->next->next =
    newNode(7);
    head->next->next->next->next->next->next->next =
    newNode(8);
  
    int k = 3;
    head = deleteKthNode(head, k);
  
    displayList(head);
  
    return 0;
}

输出：

1 2 4 5 7 8

时间复杂度： O(n)

请参阅完整文章删除链表的每个第 k 个节点以获取更多详细信息！