用于旋转链表的 Javascript 程序

给定一个单链表，将链表逆时针旋转 k 个节点。其中 k 是给定的正整数。例如，如果给定的链表是 10->20->30->40->50->60，k 为 4，则链表应修改为 50->60->10->20->30- >40。假设 k 小于链表中的节点数。

方法一：
要旋转链表，我们需要将第 k 个节点的 next 更改为 NULL，将最后一个节点的 next 更改为前一个 head 节点，最后将 head 更改为第 (k+1) 个节点。所以我们需要掌握三个节点：第k个节点、第(k+1)个节点和最后一个节点。
从头开始遍历列表并在第 k 个节点处停止。存储指向第 k 个节点的指针。我们可以使用 kthNode->next 获得第 (k+1) 个节点。继续遍历直到结束并存储指向最后一个节点的指针。最后，如上所述更改指针。

下图显示了如何在代码中使用旋转函数：

Javascript

// Javascript program to rotate
// a linked list
 
// Head of list
var head;
 
// Linked list Node
class Node
{
    constructor(val)
    {
        this.data = val;
        this.next = null;
    }
}
 
// This function rotates a linked list
// counter-clockwise and updates the head.
// The function assumes that k is smaller
// than size of linked list. It doesn't
// modify the list if k is greater than
// or equal to size
function rotate(k)
{
    if (k == 0)
        return;
 
    // Let us understand the below code
    // for example k = 4 and list =
    // 10->20->30->40->50->60.
    var current = head;
 
    // current will either point to kth or
    // NULL after this loop. current will
    // point to node 40 in the above example
    var count = 1;
 
    while (count < k && current != null)
    {
        current = current.next;
        count++;
    }
 
    // If current is NULL, k is greater than
    // or equal to count of nodes in linked list.
    // Don't change the list in this case
    if (current == null)
        return;
 
    // current points to kth node. Store it in
    // a variable. kthNode points to node 40
    // in the above example
    var kthNode = current;
 
    // current will point to last node after
    // this loop current will point to node
    // 60 in the above example
    while (current.next != null)
        current = current.next;
 
    // Change next of last node to previous
    // head Next of 60 is now changed to
    // node 10
    current.next = head;
 
    // Change head to (k+1)th node
    // head is now changed to node 50
    head = kthNode.next;
 
    // change next of kth node to null
    kthNode.next = null;
}
 
/* Given a reference (pointer to pointer) to
   the head of a list and an int, push
   a new node on the front of the list. */
    function push(new_data) {
 
/* 1 & 2: Allocate the Node &
          Put in the data */
var 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;
}
 
function printList()
{
    var temp = head;
    while (temp != null)
    {
        document.write(temp.data + " ");
        temp = temp.next;
    }
    document.write("<br/>");
}
 
// Driver code
// Create a list
// 10->20->30->40->50->60
for (i = 60; i >= 10; i -= 10)
    push(i);
 
document.write("Given list<br/>");
printList();
 
rotate(4);
 
document.write("Rotated Linked List<br/>");
printList();
// This code is contributed by todaysgaurav

输出：

Given linked list
10  20  30  40  50  60
Rotated Linked list
50  60  10  20  30  40

方法二：
要将链表旋转k，我们可以先使链表循环，然后从头节点向前移动k-1步，使第(k-1)个节点的next为空，并以第k个节点为头。

Javascript

// Javascript program to rotate
// a linked list counter clock wise
 
// Link list node
class Node
{
    constructor()
    {
        this.data = 0;
        this.next = null;
    }
}
 
var head = null;
 
// This function rotates a linked list
// counter-clockwise and updates the
// head. The function assumes that k is
// smaller than size of linked list.
function rotate(k)
{
    if (k == 0)
        return;
 
    // Let us understand the below
    // code for example k = 4 and
    // list = 10.20.30.40.50.60.
    var current = head;
 
    // Traverse till the end.
    while (current.next != null)
        current = current.next;
 
    current.next = head;
    current = head;
 
    // Traverse the linked list to
    // k-1 position which will be
    // last element for rotated array.
    for (i = 0; i < k - 1; i++)
        current = current.next;
 
    // update the head_ref and last
    // element pointer to null
    head = current.next;
    current.next = null;
}
 
// UTILITY FUNCTIONS
// Function to push a node
function push(new_data)
{
    // Allocate node
    var new_node = new Node();
 
    // Put in the data
    new_node.data = new_data;
 
    // Link the old list off the
    // new node
    new_node.next = head;
 
    // Move the head to point to the
    // new node
    head = new_node;
}
 
// Function to print linked list
function printList( node)
{
    while (node != null)
    {
        document.write(node.data + " ");
        node = node.next;
    }
}
 
// Driver code
// Start with the empty list
// Create a list 10.20.30.40.50.60
for (i = 60; i > 0; i -= 10)
    push(i);
 
document.write("Given linked list <br/>");
printList(head);
rotate(4);
 
document.write("<br/>Rotated Linked list <br/>");
printList(head);
// This code is contributed by aashish1995

输出：

Given linked list 
10 20 30 40 50 60 
Rotated Linked list 
50 60 10 20 30 40

有关详细信息，请参阅有关旋转链接列表的完整文章！