Leetcode: Palindrome Linked List

本文介绍了一种在O(n)时间内使用O(1)额外空间判断单链表是否为回文的方法。通过将链表后半部分反转并与前半部分进行比较,实现了高效的回文检测。

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Question

Given a singly linked list, determine if it is a palindrome.

Follow up:
Could you do it in O(n) time and O(1) space?

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Solution

# 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
        """

        if head==None or head.next==None:
            return True

        # segmentate list
        slow, fast = (head,)*2
        while fast.next!=None and fast.next.next!=None:
            slow = slow.next
            fast = fast.next.next
        slow = slow.next    

        head1,head2 = head,slow
        head2 = self.reverselist(head2)

        while head2!=None:
            if head1.val!=head2.val:
                return False

            head1, head2 = head1.next, head2.next

        return True

    def reverselist(self, head):
        if head==None or head.next==None:
            return head

        tail = head.next
        newhead = self.reverselist(tail)

        tail.next = head
        head.next = None

        return newhead

Error Path

  1. if head==None or head.next==None: return head should return True
  2. for segmentate list, should add slow=slow.next for the case len(list)==2
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