Palindrome Linked List

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?

1、朴素方法

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
 struct ListNode* reverseList(struct ListNode* head) {
    if(head == NULL)
        return NULL;
    struct ListNode *nex = head->next, *pre = NULL;
    while(nex != NULL)
    {
        head->next = pre;
        pre = head;
        head = nex;
        nex = nex->next;
    }
    head->next = pre;
    return head;
}
bool isPalindrome(struct ListNode* head) {
    struct ListNode *tail = head;
    int n = 0,tmp = 0;
    if(head == NULL)
        return 1;
    while(tail != NULL)
    {
        tail = tail->next;
        n++;
    }
    for(int i = 0; i < n/2; i++)    //通过不断的翻转链表来对比头尾的值
    {
        tmp = head->val;
        head = head->next;
        tail = reverseList(head);
        if(tmp != tail->val)
            return 0;
        tail = tail->next;
        head = reverseList(tail);
    }
    return 1;
}

2、网上答案：把后半段翻转，然后逐个对比值

bool isPalindrome(struct ListNode* head) {
     //如果链表为空或者仅有一个元素那么肯定是回文链表  
    if (!head || !head->next) {  
        return true;  
    }  
    //快慢指针法，寻找链表中心  
    struct ListNode * slow, *fast;  
    slow = fast = head;  
    while (fast && fast->next) {          //先检测fast是否为空，如果为空则fast->next不再运行，也就不会越界
        slow = slow->next;          //慢指针每次前进一格
        fast = fast->next->next;          //快指针每次前进两格
    }  
    if (fast) {  
        slow = reverseList(slow->next);      //如果fast不为空，则链表元素奇数个  
    }else{  
       slow = reverseList(slow);           //如果fast为空，则链表元素偶数个  
    }  
    while (slow) {  
        if (head->val != slow->val) {  
            return false;  
        }  
        slow = slow->next;  
        head = head->next;  
    }  
    return true;  
}