牛客题解 | 链表的回文结构

题目

题目链接

解题思路

这是一个判断链表是否为回文结构的问题。为了满足 \(\mathcal{O}(1)\) 的空间复杂度，我们可以：

1. 找到链表中点
2. 反转后半部分
3. 比较前后两部分
4. 恢复链表结构（可选）

关键点：

1. 使用快慢指针找中点
2. 原地反转后半部分链表
3. 比较两部分是否相同
4. 不使用额外空间

算法步骤：

1. 使用快慢指针找到中点
2. 反转后半部分链表
3. 从头和中点开始比较
4. 可选：恢复链表原始结构

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代码

class PalindromeList {
public:bool chkPalindrome(ListNode* A) {if (!A || !A->next) {return true;}// 找到中点ListNode* slow = A;ListNode* fast = A;while (fast->next && fast->next->next) {slow = slow->next;fast = fast->next->next;}// 反转后半部分ListNode* secondHalf = reverseList(slow->next);// 比较两部分ListNode* firstHalf = A;ListNode* temp = secondHalf;  // 保存起点以便恢复bool result = true;while (secondHalf) {if (firstHalf->val != secondHalf->val) {result = false;break;}firstHalf = firstHalf->next;secondHalf = secondHalf->next;}// 恢复链表结构（可选）slow->next = reverseList(temp);return result;}private:ListNode* reverseList(ListNode* head) {ListNode* prev = nullptr;ListNode* curr = head;while (curr) {ListNode* next = curr->next;curr->next = prev;prev = curr;curr = next;}return prev;}
};

import java.util.*;public class PalindromeList {public boolean chkPalindrome(ListNode A) {if (A == null || A.next == null) {return true;}// 找到中点ListNode slow = A;ListNode fast = A;while (fast.next != null && fast.next.next != null) {slow = slow.next;fast = fast.next.next;}// 反转后半部分ListNode secondHalf = reverseList(slow.next);// 比较两部分ListNode firstHalf = A;ListNode temp = secondHalf; // 保存起点以便恢复boolean result = true;while (secondHalf != null) {if (firstHalf.val != secondHalf.val) {result = false;break;}firstHalf = firstHalf.next;secondHalf = secondHalf.next;}// 恢复链表结构（可选）slow.next = reverseList(temp);return result;}private ListNode reverseList(ListNode head) {ListNode prev = null;ListNode curr = head;while (curr != null) {ListNode next = curr.next;curr.next = prev;prev = curr;curr = next;}return prev;}
}

#!/usr/bin/env python
# -*- coding: utf-8 -*-class PalindromeList:def chkPalindrome(self, A):# 处理空链表和单节点链表if not A or not A.next:return True# 找到中点slow = fast = Awhile fast.next and fast.next.next:slow = slow.nextfast = fast.next.next# 反转后半部分second_half = self.reverseList(slow.next)# 比较两部分first_half = Atemp = second_half  # 保存起点以便恢复result = Truewhile second_half:if first_half.val != second_half.val:result = Falsebreakfirst_half = first_half.nextsecond_half = second_half.next# 恢复链表结构（可选）slow.next = self.reverseList(temp)return resultdef reverseList(self, head):prev = Nonecurr = headwhile curr:next_node = curr.nextcurr.next = prevprev = currcurr = next_nodereturn prev# 测试代码
def test():# 创建测试链表: 1->2->2->1class ListNode:def __init__(self, val=0, next=None):self.val = valself.next = nexthead = ListNode(1)head.next = ListNode(2)head.next.next = ListNode(2)head.next.next.next = ListNode(1)solution = PalindromeList()print("Is palindrome:", solution.chkPalindrome(head))if __name__ == "__main__":test()

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算法及复杂度

• 算法：快慢指针 + 链表反转
• 时间复杂度：\(\mathcal{O(n)}\)
• 空间复杂度：\(\mathcal{O(1)}\)