用于查找给定链接列表的中间元素的 Javascript 程序

给定一个单链表，找到链表的中间。例如，如果给定的链表是 1->2->3->4->5，那么输出应该是 3。
如果有偶数节点，那么就会有两个中间节点，我们需要打印第二个中间元素。例如，如果给定的链表是 1->2->3->4->5->6，那么输出应该是 4。

方法一：
遍历整个链表并计算编号。的节点。现在再次遍历列表直到 count/2 并返回 count/2 处的节点。

方法二：
使用两个指针遍历链表。将一个指针移动一格，将其他指针移动二格。当快指针到达末尾时，慢指针将到达链表的中间。

下图显示了 printMiddle函数在代码中的工作方式：

Javascript

 
// JavaScript program to find
// middle of linked list
 
// Head of linked list
var head;
 
// Linked list node
class Node
{
    constructor(val)
    {
        this.data = val;
        this.next = null;
    }
}
 
// Function to print middle of
// linked list
function printMiddle()
{
    var slow_ptr = head;
    var fast_ptr = head;
    if (head != null)
    {
        while (fast_ptr != null &&
               fast_ptr.next != null)
        {
            fast_ptr = fast_ptr.next.next;
            slow_ptr = slow_ptr.next;
        }
        document.write("The middle element is [" +
                        slow_ptr.data + "] <br/><br/>");
    }
}
 
// Inserts a new Node at 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;
}
 
// This function prints contents of linked
// list starting from the given node
function printList()
{
    var tnode = head;
    while (tnode != null)
    {
        document.write(tnode.data + "->");
        tnode = tnode.next;
    }
    document.write("NULL<br/>");
}
 
for (i = 5; i > 0; --i)
{
    push(i);
    printList();
    printMiddle();
}
// This code is contributed by todaysgaurav

输出：

5->NULL
The middle element is [5]

4->5->NULL
The middle element is [5]

3->4->5->NULL
The middle element is [4]

2->3->4->5->NULL
The middle element is [4]

1->2->3->4->5->NULL
The middle element is [3]

方法三：
将 mid 元素初始化为 head 并将计数器初始化为 0。从 head 遍历列表，同时遍历递增计数器并在计数器为奇数时将 mid 更改为 mid->next。所以中间只会移动列表总长度的一半。
感谢 Narendra Kangralkar 提出这种方法。

Javascript

 
// Javascript program to implement
// the above approach
var head = null;
 
// Link list node
class Node
{
    constructor(next, val)
    {
        this.data = val;
        this.next = next;
    }
}
 
// Function to get the middle of
// the linked list
function printMiddle(head)
{
    var count = 0;
    var mid = head;
 
    while (head != null)
    {
        // Update mid, when 'count'
        // is odd number
        if ((count % 2) == 1)
            mid = mid.next;
 
        ++count;
        head = head.next;
    }
 
    // If empty list is provided
    if (mid != null)
        document.write("The middle element is [" +
                        mid.data + "]<br/><br/>");
}
 
function push(head_ref, new_data)
{
    // Allocate node
    var new_node = new Node(head_ref,
                            new_data);
 
    // Move the head to point to the new node
    head = new_node;
    return head;
}
 
// A utility function to print
// given linked list
function printList(head)
{
    while (head != null)
    {
        document.write(head.data + "-> ");
        head = head.next;
    }
    document.write("null<br/>");
}
 
// Driver code
for (i = 5; i > 0; i--)
{
    head = push(head, i);
    printList(head);
    printMiddle(head);
}
// This code contributed by gauravrajput1

输出：

5->NULL
The middle element is [5]

4->5->NULL
The middle element is [5]

3->4->5->NULL
The middle element is [4]

2->3->4->5->NULL
The middle element is [4]

1->2->3->4->5->NULL
The middle element is [3]

有关详细信息，请参阅有关查找给定链接列表的中间的完整文章！