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?
判断单链表是不是回文串，首先写了一个反转链表函数reverseList，为了省时间在isPalindrome函数里面用双指针一快一慢让慢的走到中间的结点，然后只需要反转后面的一半结点，再跟前面的一半比较即可知道是不是回文串。

struct ListNode* reverseList(struct ListNode* head) {
    if(!head)   return NULL;
    struct ListNode *prev=NULL,*next;
    while(head){
        next=head->next;
        head->next=prev;
        prev=head;
        head=next;
    }
    return prev;
}
bool isPalindrome(struct ListNode* head) {
    if(!head || !head->next) return true;
    
    struct ListNode *fast=head,*slow=head;
    //如果fast->next或者fast->next->next少一个判断都会导致慢指针走多一步导致结果错误
    while(fast && fast->next && fast->next->next){
        slow=slow->next;
        fast=fast->next->next;
    }
    struct ListNode *mid=reverseList(slow->next);
    slow->next=NULL;
    while(mid){
        if(mid->val!=head->val)
            return false;
        mid=mid->next;
        head=head->next;
    }
    return true;
}