[LeetCode] Palindrome Linked List

最新推荐文章于 2025-08-22 15:10:27 发布

转载最新推荐文章于 2025-08-22 15:10:27 发布 · 42 阅读

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原文链接：http://www.cnblogs.com/jcliBlogger/p/4635516.html

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#开发工具

本文介绍了一种通过反转单链表的后半部分并与前半部分进行比较的方法来判断链表是否为回文链表的算法实现。包括链表结构定义、反转链表函数及主函数逻辑。

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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 };