1. Introduction

The "Palindrome Linked List" problem is a common question in algorithm interviews, combining concepts of linked lists and palindromes. A palindrome is a sequence that reads the same backward as forward. This problem involves determining if the values in a singly linked list form a palindrome. It tests one's ability to manipulate and traverse linked lists efficiently.

Problem

Given the head of a singly linked list, the task is to return true if the list is a palindrome or false otherwise.

2. Solution Steps

1. Find the middle of the linked list using the fast and slow pointer technique.

2. Reverse the second half of the list.

3. Compare the nodes of the first half and the reversed second half.

4. If all corresponding nodes are equal, the list is a palindrome.

5. Restore the list to its original order by reversing the second half again (optional).

6. Return true if a palindrome, otherwise false.

3. Code Program

``````class ListNode:
def __init__(self, val=0, next=None):
self.val = val
self.next = next

while current:
next_temp = current.next
current.next = prev
prev, current = current, next_temp
return prev

# Find the middle
while fast and fast.next:
fast = fast.next.next
slow = slow.next

# Reverse the second half
second_half = reverseList(slow)

# Compare
while second_half:
if first_half.val != second_half.val:
return False
first_half, second_half = first_half.next, second_half.next

# Restore the original list (Optional)
reverseList(second_half_copy)

return True

# Example Usage
# Creating list for demonstration: [1 -> 2 -> 2 -> 1]
head = ListNode(1, ListNode(2, ListNode(2, ListNode(1))))

# List not a palindrome: [1 -> 2]

``````

```True
False
```

Explanation:

1. ListNode Class: Represents each node in the linked list.

2. Middle of List: Fast and slow pointers are used to find the middle of the list.

3. Reverse Second Half: The second half of the list is reversed for comparison.

4. Palindrome Check: The first and second halves are compared to check if the list forms a palindrome.

5. List Restoration: Optionally, the list can be restored to its original order.

6. Efficient Solution: The approach ensures an O(n) time complexity and O(1) space complexity.

7. Result: The function returns true if the list is a palindrome, otherwise false.