用于删除链表的每个第 K 个节点的 Javascript 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。

Javascript

// Javascript program to delete every 
// k-th Node of a singly linked list.
// Linked list Node 
class Node 
{
    constructor() 
    {
        this.data = 0;
        this.next = null;
    }
}
  
// To remove complete list (Needed 
// for case when k is 1)
function freeList( node) 
{
    while (node != null) 
    {
        next = node.next;
        node = next;
    }
    return node;
}
  
// Deletes every k-th node and
// returns head of modified list.
function deleteKthNode(head, k) 
{
    // If linked list is empty
    if (head == null)
        return null;
  
    if (k == 1) 
    {
        head = freeList(head);
        return null;
    }
  
    // Initialize ptr and prev before
    // starting traversal.
    var ptr = head, prev = null;
  
    // Traverse list and delete
    // every k-th node
    var 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
            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 
function displayList( head) 
{
    temp = head;
    while (temp != null) 
    {
        document.write(temp.data + 
                       " ");
        temp = temp.next;
    }
}
  
// Utility function to create a 
// new node.
function newNode(x) 
{
    temp = new Node();
    temp.data = x;
    temp.next = null;
    return temp;
}
  
// Driver Code
// Start with the empty list 
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);
  
var k = 3;
head = deleteKthNode(head, k);
displayList(head);
// This code is contributed by umadevi9616

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！