C语言每日一题（34）链表的回文结构

1->2->2->1

返回：true

class PalindromeList {
  public:
    struct ListNode* middleNode(struct ListNode* head) {//取中间结点
        struct ListNode* slow = head;
        struct ListNode* fast = head;
        while (fast && fast->next) {
            slow = slow->next;
            fast = fast->next->next;
        }
        return slow;

    }
    struct ListNode* ReverseList(struct ListNode* head ) {//逆置链表
        struct ListNode* pre = NULL;
        struct ListNode* cur = head;
        struct ListNode* nex = cur->next;
        while (cur != NULL) {
            cur->next = pre;
            pre = cur;
            cur = nex;
            nex = cur->next;
        }
        return pre;
    }
    bool chkPalindrome(ListNode* A) {//进行判断
        ListNode* middle=middleNode(A);
        ListNode* rev=ReverseList(middle);
        ListNode* cur=A;
        while(cur&&rev)
        {
            if(cur->val!=rev->val)
            {
                return false;
            }
            cur=cur->next;
            rev=rev->next;
        }
        return true;
    }
};