用于将仲裁指针指向链接列表中的最大值右侧节点的 Javascript 程序

给定单链表，每个节点都有一个当前指向 NULL 的附加“任意”指针。我们需要将“任意”指针指向右侧链表中的最大值节点。

一个简单的解决方案是一个一个地遍历所有节点。对于每个节点，找到右侧具有最大值的节点并更改下一个指针。该解决方案的时间复杂度为 O(n ² )。

一个有效的解决方案可以在 O(n) 时间内工作。以下是步骤。

1. 反转给定的链表。
2. 开始遍历链表，存储目前遇到的最大值节点。使每个节点的仲裁指向最大值。如果当前节点中的数据超过了迄今为止的max节点，则更新max。
3. 反向修改链表并返回头部。

以下是上述步骤的实现。

Javascript

// Javascript program to point arbit pointers 
// to highest value on its right 
  
// Link list node 
class Node 
{
    constructor(val) 
    {
        this.data = val;
        this.arbit = null;
        this.next = null;
    }
}
      
// Function to reverse the linked list 
function reverse(head) 
{
    var prev = null, 
        current = head, 
        next = null;
    while (current != null) 
    {
        next = current.next;
        current.next = prev;
        prev = current;
        current = next;
    }
    return prev;
}
  
// This function populates arbit pointer 
// in every node to the greatest value 
// to its right.
function populateArbit(head) 
{
    // Reverse given linked list
    head = reverse(head);
  
    // Initialize pointer to maximum 
    // value node
    var max = head;
  
    // Traverse the reversed list
    var temp = head.next;
    while (temp != null) 
    {
        // Connect max through 
        // arbit pointer
        temp.arbit = max;
  
        // Update max if required
        if (max.data < temp.data)
            max = temp;
  
        // Move ahead in reversed list
        temp = temp.next;
    }
  
    // Reverse modified linked list 
    // and return head.
    return reverse(head);
}
  
// Utility function to print result 
// linked list
function printNextArbitPointers(node) 
{
    document.write("Node     " + 
                   "Next Pointer    " + 
                   "Arbit Pointer<br/>");
    while (node != null) 
    {
        document.write(node.data + 
                       "          ");
        if (node.next != null)
            document.write(node.next.data + 
                           "                    ");
         else
             document.write("NULL" + 
                            "          ");
  
         if (node.arbit != null)
             document.write(node.arbit.data);
         else
             document.write("NULL");
  
         document.write("<br/>");
         node = node.next;
    }
}
  
/* Function to create a new node 
   with given data */
function newNode(data) 
{
    var new_node = new Node();
    new_node.data = data;
    new_node.next = null;
    return new_node;
}
  
// Driver code 
var head = newNode(5);
head.next = newNode(10);
head.next.next = newNode(2);
head.next.next.next = newNode(3);
head = populateArbit(head);
document.write(
         "Resultant Linked List is: <br/>");
printNextArbitPointers(head);
// This code is contributed by umadevi9616 

输出：

Resultant Linked List is: 
Node    Next Pointer    Arbit Pointer
5               10              10
10              2               3
2               3               3
3               NULL            NULL

递归解决方案：
我们可以递归到达最后一个节点，从最后遍历链表。递归解决方案不需要反转链表。我们还可以使用堆栈代替递归来临时保存节点。感谢 Santosh Kumar Mishra 提供此解决方案。

Javascript

// Javascript program to point arbit pointers 
// to highest value on its right
  
// Link list node 
class Node 
{
    constructor() 
    {
         this.data = 0;
         this.arbit = null;
         this.next = null;
    }
}
  
var maxNode;
  
// This function populates arbit pointer 
// in every node to the greatest value 
// to its right.
function populateArbit(head) 
{
    // If head is null simply return 
    // the list
    if (head == null)
        return;
  
    /* if head->next is null it means we 
       reached at the last node just update 
       the max and maxNode */
    if (head.next == null) 
    {
        maxNode = head;
        return;
    }
  
    /* Calling the populateArbit to the
       next node */
    populateArbit(head.next);
  
    /* Updating the arbit node of the current
       node with the maximum value on the
       right side */
    head.arbit = maxNode;
  
    /* if current Node value id greater then 
       the previous right node then update it */
    if (head.data > maxNode.data)
        maxNode = head;
  
    return;
}
  
// Utility function to print result 
// linked list
function printNextArbitPointers(node) 
{
    document.write(
             "Node Next Pointer Arbit Pointer<br/>");
    while (node != null) 
    {
        document.write(node.data + 
                       "     ");
  
        if (node.next != null)
            document.write(node.next.data + 
                           "     ");
         else
             document.write("NULL" + 
                            "   ");
  
         if (node.arbit != null)
             document.write(node.arbit.data);
         else
             document.write("NULL");
  
         document.write("<br/>");
         node = node.next;
    }
}
  
/* Function to create a new node 
   with given data */
function newNode(data) 
{
    var new_node = new Node();
    new_node.data = data;
    new_node.next = null;
    return new_node;
}
  
// Driver code 
var head = newNode(5);
head.next = newNode(10);
head.next.next = newNode(2);
head.next.next.next = newNode(3);
populateArbit(head);
document.write(
         "Resultant Linked List is: <br/>");
printNextArbitPointers(head);
// This code contributed by gauravrajput1

输出：

Resultant Linked List is: 
Node    Next Pointer    Arbit Pointer
5               10              10
10              2               3
2               3               3
3               NULL            NULL

请参考完整文章 Point arbit pointer to maximum value right side node in alinked list 了解更多详情！