Leetcode 234. Palindrome Linked List

题目链接：https://leetcode.com/problems/palindrome-linked-list/

题目描述：

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?

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思路：根据题目要求，时间复杂度O(n)，空间复杂度O(1),采用反转链表的方法，将前半部分反转，然后挨个比较。

代码：

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:bool isPalindrome(ListNode* head) {bool re = 1;if (head == NULL || head->next == NULL)re = 1;else{ListNode* p1 = head;//慢指针ListNode* p2 = head->next;ListNode* p3 = head->next;//p3是快指针，移动速度是p1的两倍p1->next = NULL;//反转前半部分链表while (p3!= NULL&&p3->next != NULL){p3 = p3->next->next;ListNode* tmp = p2->next;p2->next = p1;p1 = p2;p2 = tmp;}if (p3 == NULL)//链表长度是奇数，p1指向了中间结点，让p1指向它的前一个结点p1 = p1->next;//开始比较while (p2!= NULL){if (p1->val != p2->val){re = 0;break;}p1 = p1->next;p2 = p2->next;}}return re;}};