1、用快慢指標找到連結串列中心位置
如果連結串列長度為偶數個，比如1,2,3,4，最後找到3
如果連結串列長度為奇數個，比如1,2,3,4,5，最好找到3
2、將後半段反轉
3、從一頭一尾狀態開始判斷是否為迴文
c++程式碼如下

/**
 * 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) return true;

        if(head->next == NULL) return true;

        if(head->next->next == NULL) {

            if(head->val == head->next->val) return true;
            else return false;

        } 

        ListNode *
 
pSlow, *pFast;
        pSlow = head;
        pFast = head;
        while(pFast->next != NULL) {

            pFast = pFast->next;
            if(pFast->next) pFast = pFast->next;

            pSlow = pSlow->next;

        }

        ListNode *p1 = pSlow;
        ListNode *p2 = p1->
 
next;
        ListNode *p3 = p2->next;
        while(true) {

            p2->next = p1;

            p1 = p2;
            p2 = p3;
            if(p3 == NULL) 

                break;

            else 

                p3 = p3->next;

        }

        pSlow->next = NULL;

        ListNode *p4;
        ListNode *p5;
        p4 = head;
        p5 = pFast;
        while(p5 != NULL) {

            if(p4->val != p5->val) return false;

            p4 = p4->next;
            p5 = p5->next;

        }

        return true;

    }
};