234.md

Palindrome Linked List

Description

link

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Solution

• Find the mid node of linked list
• Reverse the second half part, (slow is exactly mid one or in the first half part)
• Compare two parts

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Code

T : O(n) S : O(1)

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def isPalindrome(self, head):
        """
        :type head: ListNode
        :rtype: bool
        """
        # find the mid node
        slow, fast = head, head
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        # reverse the second-half
        node = None
        while slow:
            nxt = slow.next
            slow.next = node
            node = slow
            slow = nxt
        # compare two part
        while node # while node and head: (why head no need?)
            if node.val != head.val:
                return False
            node = node.next
            head = head.next
            
        return True