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?
判断一个链表是否回文？要求O（n）时间复杂度，O（1）空间复杂度。
快慢指针找到中点，将中点以后的逆置，再逐个判断是否相等
代码语言：javascript复制/**
* 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 || !head->next) return true;
ListNode* fast = head;
ListNode* slow = head;
bool flag;
while(fast->next)
{
flag = true;
fast = fast->next;
if(fast->next)
{
flag = false;
slow = slow->next;
fast = fast->next;
}
}
fast = slow->next;
slow->next = NULL;
if(flag) slow = slow->next;
while(fast)
{
ListNode* temp = fast->next;
fast->next = slow;
slow = fast;
fast = temp;
}
while(slow && head)
{
if(slow->val != head->val) return false;
slow = slow->next;
head = head->next;
}
if(slow || head) return false;
return true;
}
};
