[LeetCode] Palindrome Linked List

最新推荐文章于 2025-09-05 14:41:09 发布

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最新推荐文章于 2025-09-05 14:41:09 发布

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文章标签：开发工具

原文链接：https://yq.aliyun.com/articles/331463

本文介绍了一种通过反转链表右半部分并比较其与左半部分来判断链表是否为回文的方法。通过实现反转链表的函数，我们能够有效地检查链表的对称性。

The idea is not so obvious at first glance. Since you cannot move from a node back to its previous node in a singly linked list, we choose to reverse the right half of the list and then compare it with the left half. The code is as follows. It shoul be obvious after you run some examples.

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11     bool isPalindrome(ListNode* head) {
12         if (!head || !(head -> next)) return true;
13         ListNode* slow = head;
14         ListNode* fast = head;
15         while (fast && fast -> next) {
16             slow = slow -> next;
17             fast = fast -> next -> next;
18         }
19         if (fast) {
20             slow -> next = reverseList(slow -> next);
21             slow = slow -> next;
22         }
23         else slow = reverseList(slow);
24         while (slow) {
25             if (head -> val != slow -> val)
26                 return false;
27             head = head -> next;
28             slow = slow -> next;
29         }
30         return true;
31     }
32 private:
33     ListNode* reverseList(ListNode* node) {
34         ListNode* pre = NULL;
35         while (node) {
36             ListNode* next = node -> next;
37             node -> next = pre;
38             pre = node;
39             node = next;
40         }
41         return pre;
42     }
43 };