LeetCode 234. Palindrome Linked List

Given a singly linked list, determine if it is a palindrome.

Example 1:

Input: 1->2
Output: false

Example 2:

Input: 1->2->2->1
Output: true

Follow up:
Could you do it in O(n) time and O(1) space?

解析：

判断一个链表是不是回文链表，即链表首尾对称。并且follow up中指出空间复杂度为O(n)，时间复杂度为O(1)。考虑首先找到链表的中间节点，可以通过快慢指针的方式。然后将链表后半段逆序。然后再比较两个链表对应元素是否相同。代码如下：

/**
 * 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) {
       if(head == NULL || head->next == NULL)
           return true;
        ListNode *slow =head, *fast = head;
        while(fast->next && fast->next->next)//快慢指针找到链表中间节点
        {
            slow = slow->next;
            fast = fast->next->next;
        }
        ListNode* pre = NULL, *next = NULL;
        ListNode* temp = slow->next;
        while(temp)//后半部分链表逆序
        {
            next = temp->next;
            temp->next = pre;
            pre = temp;
            temp = next;
        }
        while(pre)//比较两个链表对应元素是否相等
        {
            if(pre->val != head->val)
                return false;
            pre=pre->next;
            head = head->next;
        }
        return true;
    }
};