14. 判断单链表是否为回文链表

#1

RT

#2

hi，可以不用先遍历一次链表计算len，增加一个快指针，如下：

bool isPalindrome(ListNode* head) {
        if (head == NULL) {
            return true;
        }
        if (head->next == NULL) {
            return true;
        }

        ListNode* p = head;
        ListNode* cur = head->next;
        ListNode* fast = head->next->next;
        while(fast != NULL && fast->next != NULL) {
            p->next = cur->next;
            cur->next = head;
            head = cur;
            cur = p->next;

            fast = fast->next->next;
        }

        if (fast != NULL) {
            cur = cur->next;
        }
        ListNode* p1 = head;
        while(cur != NULL) {
            if (p1->val != cur->val) {
                return false;
            }
            p1 = p1->next;
            cur = cur->next;
        }

        return true;
    }