Given a linked list with a loop, the task is to find whether it is palindrome or not. You are not allowed to remove the loop.
Examples:
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.
Algorithm:
- Detect the loop using the Floyd Cycle Detection Algorithm.
- Then find the starting node of the loop as discussed in this.
- Check the linked list is palindrome or not as discussed in this.
Below is the implementation.
Javascript
<script> // 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 </script> |
Output:
Palindrome
Time Complexity: O(n) where n is no of nodes in the linked list
Auxiliary Space: O(n) where n is size of List
Please refer complete article on Check linked list with a loop is palindrome or not for more details!