# 1. Introduction

The "Palindrome Linked List" problem is an interesting challenge that tests one's ability to manipulate linked lists. A palindrome is a sequence that reads the same forward and backward, and this problem requires checking if a singly linked list forms a palindrome. It combines concepts of linked list traversal, reversal, and comparison.

## Problem

Given the head of a singly linked list, the objective is to determine whether the list forms a palindrome. The function should return true if the list is a palindrome, and false otherwise.

# 2. Solution Steps

1. Traverse the list to find its length.

2. Identify the midpoint of the list.

3. Reverse the second half of the list.

4. Compare the first half and the reversed second half node by node.

5. If all corresponding nodes match, the list is a palindrome.

6. Optionally, reverse the second half again to restore the list's original order.

7. Return true if a palindrome is detected, otherwise false.

# 3. Code Program

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

return True

# Find the length of the list
while current:
length += 1
current = current.next

# Find the midpoint and reverse the second half
mid = length // 2
for _ in range(mid):
prev, current = current, current.next
second_half, prev.next = prev, None

# Reverse the second half
prev, current = None, current
while current:
next_node = current.next
current.next = prev
prev, current = current, next_node
second_half_reversed = prev

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

return True

# Example Usage
head = ListNode(1, ListNode(2, ListNode(2, ListNode(1))))

``````

```True
False
```

### Explanation:

1. ListNode Class: Represents nodes in the linked list.

2. List Length Calculation: Traverses the list to find its total length.

3. Midpoint Identification: Determines the middle of the list for division.

4. Reversal of Second Half: Reverses the second half of the list for comparison.

5. Palindrome Check: Compares the first half and the reversed second half node by node.

6. Restoration (Optional): Restores the second half to its original order.

7. Efficient Algorithm: Provides an O(n) time complexity solution with O(1) space complexity.

8. Result: Returns true if the list is a palindrome, otherwise false.