【VIP】【链表翻转】234. Palindrome Linked List【E】【94】

最新推荐文章于 2025-12-06 07:43:37 发布

原创最新推荐文章于 2025-12-06 07:43:37 发布 · 442 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode题解 #leetcode #链表 #双指针

Python 同时被 3 个专栏收录

176 篇文章

订阅专栏

leetcode

175 篇文章

订阅专栏

算法

157 篇文章

订阅专栏

本文介绍了一种在O(n)时间和O(1)空间复杂度下判断单链表是否为回文的有效方法。通过设置快慢指针找到链表中点，并在遍历过程中反转前半部分链表，最后比较前后两部分实现回文验证。

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?

Subscribe to see which companies asked this question

两种方法，注释掉的比较流氓，把链表的东西拿出来放进list里，然后翻转比较

第二种方法比较正常，而且也满足O(1)的空间条件，设置一快一慢两个pointer，快的走到头的时候，慢的正好在中间

慢的每走一步，都进行一次链表翻转。

然后逐个节点比较

# 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):
        if head == None or head.next == None :
            return True
        
        h = head 
        fast = head
        slow = head
        prv = None
        temp = None
        
        while fast and fast.next:
            fast = fast.next.next
            temp = slow.next
            slow.next = prv
            prv = slow
            slow = temp
        
        slow = prv
        if fast:
            temp = temp.next
        
        '''
        while temp:
            print temp.val
            temp = temp.next
        print '----'
        while slow:
            print slow.val
            slow = slow.next
        '''
        
        while temp and slow:
            #print temp.val
            if temp.val != slow.val:
                return False
            temp = temp.next
            slow = slow.next
        return True
        
            
        
        '''
        q = []
        
        h = head
        while h :
            q.append(h.val)
            h = h.next
        qq = q[:]
        
        qq.reverse()
        return q == qq
        '''
        
        
        """
        :type head: ListNode
        :rtype: bool
        """