Given a singly linked list, determine if it is a palindrome.
Follow up:
Could you do it in O(n) time and O(1) space?
思路：将链表前半部分反转，对比就行了。
1 class Solution { 2 public: 3 bool isPalindrome(ListNode* head) { 4 if (head == NULL || head->next == NULL) 5 return true; 6 int count = 1; 7 ListNode prenode(-1); 8 prenode.next = head; 9 ListNode *cur = head, *nex, *pre = &prenode; 10 while (cur = cur->next) 11 count++; 12 cur = head; 13 nex = cur->next; 14 for (int i = 1; i < count / 2; i++) 15 { 16 cur->next = nex->next; 17 nex->next = pre->next; 18 pre->next = nex; 19 nex = cur->next; 20 } 21 if (count % 2) nex = nex->next; 22 cur = pre->next; 23 for (int i = 1; i <= count / 2; i++, cur = cur->next, nex = nex->next) 24 if (cur->val != nex->val) return false; 25 return true; 26 } 27 };