用循环检查链表是否为回文的 Javascript 程序

给定一个带有循环的链表，任务是找出它是否是回文。不允许删除循环。

例子：

Input: 1 -> 2 -> 3 -> 2
             /|      |/
               ------- 1  
Output: Palindrome
Linked list is 1 2 3 2 1 which is a 
palindrome.

Input: 1 -> 2 -> 3 -> 4
             /|      |/
               ------- 1  
Output: Not Palindrome
Linked list is 1 2 3 4 1 which is a 
not palindrome. 

算法：

1. 使用弗洛伊德循环检测算法检测环路。
2. 然后找到循环的起始节点，如本文所述。
3. 检查链表是否为回文，如本文所述。

下面是实现。

Javascript

// Javascript program to check if a 
// linked list with loop is palindrome 
// or not.
class GfG{
  
// Link list node 
class Node 
{
    constructor() 
    {
        this.data = 0;
        this.next = null;
    }
}
  
/* Function to find loop starting node. 
   loop_node --> Pointer to one of the loop
   nodes head --> Pointer to the start node 
   of the linked list */
function getLoopstart(loop_node, head) 
{
    var ptr1 = loop_node;
    var ptr2 = loop_node;
  
    // Count the number of nodes in 
    // loop
    var k = 1, i;
    while (ptr1.next != ptr2) 
    {
        ptr1 = ptr1.next;
        k++;
    }
  
    // Fix one pointer to head
    ptr1 = head;
  
    // And the other pointer to k
    // nodes after head
    ptr2 = head;
    for (i = 0; i < k; i++)
        ptr2 = ptr2.next;
  
    /* Move both pointers at the same pace, 
       they will meet at loop starting node */
    while (ptr2 != ptr1) 
    {
        ptr1 = ptr1.next;
        ptr2 = ptr2.next;
    }
    return ptr1;
}
  
/* This function detects and find loop 
   starting node in the list */
function detectAndgetLoopstarting(head) 
{
    var slow_p = head, fast_p = head, 
        loop_start = null;
  
    // Start traversing list and detect loop
    while (slow_p != null && fast_p != null && 
           fast_p.next != null) 
    {
        slow_p = slow_p.next;
        fast_p = fast_p.next.next;
  
         /* If slow_p and fast_p meet then find 
            the loop starting node */
         if (slow_p == fast_p) 
         {
             loop_start = getLoopstart(slow_p, head);
             break;
         }
    }
  
    // Return starting node of loop
    return loop_start;
}
  
// Utility function to check if a linked 
// list with loop is palindrome with given 
// starting point.
function isPalindromeUtil(head, loop_start) 
{
    var ptr = head;
    var s =  [];
  
    // Traverse linked list until last node
    // is equal to loop_start and store the
    // elements till start in a stack
    var count = 0;
    while (ptr != loop_start || 
           count != 1) 
    {
        s.push(ptr.data);
        if (ptr == loop_start)
            count = 1;
         ptr = ptr.next;
    }
    ptr = head;
    count = 0;
  
    // Traverse linked list until last 
    // node is equal to loop_start second 
    // time
    while (ptr != loop_start || 
           count != 1) 
    {
        // Compare data of node with 
        // the top of stack
        // If equal then continue
        var stk = s.pop();
        if (ptr.data == stk);                
  
        // Else return false
        else
        {
            s.push(stk);
            return false;
        }
  
        if (ptr == loop_start)
                count = 1;
            ptr = ptr.next;
    }
    // Return true if linked list is 
    // palindrome
    return true;
}
      
// Function to find if linked list
// is palindrome or not
function isPalindrome(head) 
{
    // Find the loop starting node
    var loop_start = 
        detectAndgetLoopstarting(head);
  
    // Check if linked list is palindrome
    return isPalindromeUtil(head, 
                            loop_start);
}
  
function newNode(key) 
{
    var temp = new Node();
    temp.data = key;
    temp.next = null;
    return temp;
}
  
// Driver code 
var head = newNode(50);
head.next = newNode(20);
head.next.next = newNode(15);
head.next.next.next = newNode(20);
head.next.next.next.next = 
newNode(50);
  
// Create a loop for testing 
head.next.next.next.next.next = 
head.next.next;
  
if (isPalindrome(head) == true)
    document.write("Palindrome");
else
    document.write("Not Palindrome");
// This code is contributed by aashish1995 

输出：

Palindrome

请参阅完整的文章检查带有循环的链表是否回文以获取更多详细信息！