leetcode算法练习: 回文链表

题目：请判断一个链表是否为回文链表。

示例 1:

输入: 1->2
输出: false
示例 2:

输入: 1->2->2->1
输出: true
进阶：
你能否用 O(n) 时间复杂度和 O(1) 空间复杂度解决此题？

思路：
先把整个链表从中间分开，分为两段
然后把后面那个链表到过来
最后一一判断对应值是否相等
leetcode源代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */
//反转链表函数
struct ListNode* reverseList(struct ListNode* Head){
    struct ListNode *pr=NULL,*p;
    if(Head==NULL)
        return NULL;
    while (Head->next!=NULL){
        p=Head;
        Head=Head->next;
        p->next=pr;
        pr=p;
    }
    Head->next=pr;
    pr=Head;
    return pr;
}

bool isPalindrome(struct ListNode* head){
    if(head==NULL||head->next==NULL)
        return 1;
    struct ListNode *p1=head,*p2=head;
    //把链表分为两点断并找到中间靠后那一段链表的开头
    while(p2->next&&p2->next->next){
        p2=p2->next->next;
        p1=p1->next;
    }
    p1=p1->next;
 //   调用反转链表函数
    p1=reverseList(p1);
    //一一判断对应值是否相等
    while(p1){
        if(p1->val!=head->val)
            return 0;
        p1=p1->next;
        head=head->next;
    }
    return 1;
}

leetcode标准答案：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */

struct ListNode* reverList(struct ListNode* pHead)
{
    struct ListNode revertHead;
    revertHead.next = NULL;
    while (pHead) {
        struct ListNode *tmp = pHead->next;
        pHead->next = revertHead.next;
        revertHead.next = pHead;
        pHead = tmp;
    }

    return revertHead.next;
}

bool isPalindrome(struct ListNode* head){
    struct ListNode* fast = head;
    struct ListNode* slow = head;
    struct ListNode* mid = NULL;
    struct ListNode* pHead = head;
    if (head == NULL || head->next == NULL) {
        return true;
    }
    while (fast->next && fast->next->next) {
        fast = fast->next->next;
        slow = slow->next;
    }

    mid = slow->next;
    slow->next = NULL;
    mid = reverList(mid);
    while (pHead && mid) {
        if (pHead->val != mid->val) {
            return false;
        }

        pHead = pHead->next;
        mid = mid->next;
    }

    if (pHead) {
        return true;
    }

    return true;
}