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?
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * struct ListNode *next; 6 * }; 7 */ 8 bool isPalindrome(struct ListNode* head) { 9 struct ListNode* q; //P走一步，q走两部，q到尾部，p到中间节点 10 struct ListNode* p; 11 struct ListNode* head2; 12 p = head; 13 q = head; 14 if(head == NULL) //空链表也算 15 return 1; 16 while(q->next != NULL) 17 { 18 p = p->next; 19 q = q->next; 20 if(q->next == NULL) 21 break; 22 q = q->next; 23 } 24 head2 = p; 25 while(p != q) //后部分链表逆置 26 { 27 head2 = head2->next; 28 p->next = q->next; 29 q->next = p; 30 p = head2; 31 } 32 p = head; //从开头遍历2个链表 33 q = head2; 34 while(p != NULL && q != NULL) 35 { 36 if(p->val != q->val) 37 break; 38 p = p->next; 39 q = q->next; 40 41 } 42 if( NULL == p || NULL == q) //达到尾结点，即是回文链表 43 return 1; 44 return 0; 45 }