【Leetcode_总结】234. 回文链表 - python

本文介绍了一种判断链表是否为回文的方法,通过遍历链表并将节点值存入数组,再比较数组与其反转后的相等性来判断。提供了一个Python实现示例。

Q:

请判断一个链表是否为回文链表。

示例 1:

输入: 1->2
输出: false

示例 2:

输入: 1->2->2->1
输出: true

链接:https://leetcode-cn.com/problems/palindrome-linked-list/description/

思路:遍历链表,判断遍历结果是否是回文串

代码:

# 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 not head:
            return True
        tmp = []
        pre = head
        while pre.next:
            tmp.append(pre.val)
            pre = pre.next
        tmp.append(pre.val)
        return tmp == tmp[::-1]
        

以下是几种 LeetCode 234回文链表问题的 Python 实现: ### 方法一:将链表复制到数组里再从两头比对 ```python # 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: ListNode) -> bool: lst = [] node = head while node: lst.append(node.val) node = node.next start = 0 end = len(lst) - 1 while start < end: if lst[start] != lst[end]: return False start += 1 end -= 1 return True ``` ### 方法二:递归法 ```python # Definition for singly-linked list. class ListNode(object): def __init__(self, val=0, next=None): self.val = val self.next = next class Solution(object): def isPalindrome(self, head): front_pointer = head def recursively_check(current_node=head): if current_node is not None: if not recursively_check(current_node.next): return False if front_pointer.val != current_node.val: return False nonlocal front_pointer front_pointer = front_pointer.next return True return recursively_check() ``` ### 方法三:快慢指针 + 反转链表 ```python # Definition for singly-linked list. class ListNode: def __init__(self, x): self.val = x self.next = None class Solution: def isPalindrome(self, head: ListNode) -> bool: if head is None: return True first_half_end = self.end_of_first_half(head) second_half_start = self.reverse_list(first_half_end.next) result = True first_position = head second_position = second_half_start while result and second_position is not None: if first_position.val != second_position.val: result = False first_position = first_position.next second_position = second_position.next first_half_end.next = self.reverse_list(second_half_start) return result def end_of_first_half(self, head): fast = head slow = head while fast.next is not None and fast.next.next is not None: fast = fast.next.next slow = slow.next return slow def reverse_list(self, head): previous = None current = head while current is not None: next_node = current.next current.next = previous previous = current current = next_node return previous ```
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