234. Palindrome Linked List(팰린드롬 연결 리스트)

풀이 1)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        
        rev = None
        slow = fast = head
        
        while fast and fast.next:
            fast = fast.next.next
            rev, rev.next, slow = slow, rev, slow.next
            
        if fast:
            slow = slow.next
        
        while rev and rev.val == slow.val:
            slow, rev = slow.next, rev.next
        return not rev

풀이 2)

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def isPalindrome(self, head: Optional[ListNode]) -> bool:
        
        q = collections.deque()
        
        if not head:
            return True
        
        node = head
        
        while node is not None:
            q.append(node.val)
            node = node.next
        
        while len(q) > 1:
            if q.pop() != q.popleft():
                return False
        
        return True