题目： Given a singly linked list, determine if it is a palindrome.
思路： 给定一个链表，判断它是不是回文链表 根据链表的奇偶分情况，然后反转后半段链表，与前半段比较。 代码：
struct ListNode { int val; ListNode *next; ListNode(int x) : val(x), next(NULL) {} }; class Solution { public: bool isPalindrome(ListNode* head) { ListNode* slow, *fast;//利用快慢指针 slow = fast = head; while (fast && fast->next){ slow = slow->next; fast = fast->next->next; } if (fast){//判断链表的奇偶，奇数就不计中间那个节点 slow->next = reverseList(slow->next); slow = slow->next; } else{//偶数直接反转 slow = reverseList(slow); } while (slow){//判断后半段反转后与前半段是否相等 if (head->val != slow->val){ return false; } slow = slow->next; head = head->next; } return true; } ListNode* reverseList(ListNode* head) {//反转链表 ListNode* newHead = NULL; while (head) { ListNode* nextNode = head->next; head->next = newHead; newHead = head; head = nextNode; } return newHead; } };