leetcode_c++：链表：Palindrome Linked List(234)

Given a singly linked list, determine if it is a palindrome.
// 判断是否链表是回文

---

时间：O(n)
空间：O(1)

//先将一般翻转，然后两个指针向两边走

---

class Solution {public:    ListNode* findMiddle(ListNode* head){        ListNode* p1=head,*p2=head;        while(p2 &&p2->next ){            p1=p1->next;            p2=p2->next->next;        }        return p1;    }    ListNode* reverseLink(ListNode* head){        ListNode* p=NULL;        while(head){            ListNode* q=head->next;            head->next=p;            p=head;            head=q;        }        return p;    }    bool isPalindrome(ListNode* head) {        ListNode* pMid=findMiddle(head);        ListNode* pRev=reverseLink(pMid);        for(;head!=pMid;head=head->next,pRev=pRev->next){            if(head->val!=pRev->val)                return false;        }        return true;    }};