用于反转单链表中的备用 K 节点的 Javascript 程序

给定一个链表，编写一个函数以有效的方式反转每个交替的 k 个节点（其中 k 是函数的输入）。给出算法的复杂性。

例子：

Inputs:   1->2->3->4->5->6->7->8->9->NULL and k = 3
Output:   3->2->1->4->5->6->9->8->7->NULL. 

方法 1（处理 2k 个节点并递归调用列表的其余部分）：
此方法基本上是本文中讨论的方法的扩展。

kAltReverse(struct node *head, int k)
  1)  Reverse first k nodes.
  2)  In the modified list head points to the kth node.  So change next 
       of head to (k+1)th node
  3)  Move the current pointer to skip next k nodes.
  4)  Call the kAltReverse() recursively for rest of the n - 2k nodes.
  5)  Return new head of the list.

Javascript

// JavaScript program to reverse 
// alternate k nodes in a linked list
class Node 
{
    constructor(d)
    {
        this.data = d;
        this.next = null;
    }
}
  
let head;
  
// Reverses alternate k nodes and returns
// the pointer to the new head node 
function kAltReverse(node, k)
{
    let current = node;
    let next = null, prev = null;
    let count = 0;
  
    /* 1) reverse first k nodes of the 
          linked list */
    while (current != null && count < k)
    {
        next = current.next;
        current.next = prev;
        prev = current;
        current = next;
        count++;
    }
  
    /* 2) Now head points to the kth node.  
          So change next of head to 
          (k+1)th node*/
    if (node != null)
    {
        node.next = current;
    }
  
    /* 3) We do not want to reverse next k 
          nodes. So move the current pointer 
          to skip next k nodes */
    count = 0;
    while (count < k - 1 && 
           current != null) 
    {
        current = current.next;
        count++;
    }
  
    /* 4) Recursively call for the list starting
          from current->next. And make rest of 
          the list as next of first node */
    if (current != null) 
    {
        current.next = 
                kAltReverse(current.next, k);
    }
  
    /* 5) prev is new head of the 
          input list */
    return prev;
}
  
function printList(node)
{
    while (node != null)
    {
        document.write(node.data + " ");
        node = node.next;
    }
}
  
function push(newdata)
{
    let mynode = new Node(newdata);
    mynode.next = head;
    head = mynode;
}
  
// Driver code
  
// Creating the linkedlist
for(let i = 20; i > 0; i--)
{
    push(i);
}
document.write("Given Linked List :<br>");
printList(head);
head = kAltReverse(head, 3);
  
document.write("<br>");
document.write("Modified Linked List :<br>");
printList(head);
// This code is contributed by rag2127

输出：

Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19

时间复杂度： O(n)

方法2（处理k个节点并递归调用列表的其余部分）：
方法1反转第一个k个节点，然后将指针向前移动到k个节点。因此方法 1 使用两个 while 循环，并在一次递归调用中处理 2k 个节点。

此方法在递归调用中仅处理 k 个节点。它使用第三个布尔参数 b 来决定是反转 k 个元素还是简单地移动指针。

_kAltReverse(struct node *head, int k, bool b)
  1)  If b is true, then reverse first k nodes.
  2)  If b is false, then move the pointer k nodes ahead.
  3)  Call the kAltReverse() recursively for rest of the n - k nodes and link 
       rest of the modified list with end of first k nodes. 
  4)  Return new head of the list.

Javascript

// Javascript program to reverse 
// alternate k nodes in a linked list
var head;
  
class Node 
{
    constructor(val) 
    {
        this.data = val;
        this.next = null;
    }
}
       
/* Alternatively reverses the given 
   linked list in groups of given size k. */
function kAltReverse(head, k) 
{
    return _kAltReverse(head, k, true);
}
  
/* Helper function for kAltReverse(). 
   It reverses k nodes of the list only 
   if the third parameter b is passed as 
   true, otherwise moves the pointer k 
   nodes ahead and recursively calls itself */
function _kAltReverse(node, k, b) 
{
    if (node == null) 
    {
        return null;
    }
  
    var count = 1;
    var prev = null;
    var current = node;
    var next = null;
  
    /* The loop serves two purposes 
       1) If b is true, then it reverses 
          the k nodes 
       2) If b is false, then it moves the
          current pointer */
    while (current != null && count <= k) 
    {
        next = current.next;
  
        // Reverse the nodes only if b is true 
        if (b == true) 
        {
            current.next = prev;
        }
  
        prev = current;
        current = next;
        count++;
    }
  
    /* 3) If b is true, then the node is the kth 
          node. So attach the rest of the list 
          after node. 
       4) After attaching, return the new head */
    if (b == true) 
    {
        node.next = 
            _kAltReverse(current, k, !b);
        return prev;
    } 
  
    /* If b is not true, then attach rest of 
       the list after prev. So attach rest of 
       the list after prev */ 
    else 
    {
        prev.next = _kAltReverse(current, k, !b);
        return node;
    }
}
  
function printList(node) 
{
    while (node != null) 
    {
        document.write(node.data + " ");
        node = node.next;
    }
}
  
function push(newdata) 
{
    var mynode = new Node(newdata);
    mynode.next = head;
    head = mynode;
}
  
// Creating the linkedlist
for (i = 20; i > 0; i--) 
{
    push(i);
}
document.write("Given Linked List :<br/>");
printList(head);
head = kAltReverse(head, 3);
document.write("<br/>");
document.write("Modified Linked List :<br/>");
printList(head);
// This code is contributed by aashish1995

输出：

Given linked list
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Modified Linked list
3 2 1 4 5 6 9 8 7 10 11 12 15 14 13 16 17 18 20 19

时间复杂度： O(n)

有关详细信息，请参阅有关单链表中反向备用 K 节点的完整文章！