*Palindrome Linked List

 1 /**
 2  * 本代码由九章算法编辑提供。没有版权欢迎转发。
 3  * - 九章算法致力于帮助更多中国人找到好的工作，教师团队均来自硅谷和国内的一线大公司在职工程师。
 4  * - 现有的面试培训课程包括：九章算法班，系统设计班，BAT国内班
 5  * - 更多详情请见官方网站：http://www.jiuzhang.com/
 6  */
 7 
 8 // This code would destroy the original structure of the linked list.
 9 // If you do not want to destroy the structure, you can reserve the second part back.
10 public class Solution {
11     /**
12      * @param head a ListNode
13      * @return a boolean
14      */
15     public boolean isPalindrome(ListNode head) {
16         if (head == null) {
17             return true;
18         }
19         
20         ListNode middle = findMiddle(head);
21         middle.next = reverse(middle.next);
22         
23         ListNode p1 = head, p2 = middle.next;
24         while (p1 != null && p2 != null && p1.val == p2.val) {
25             p1 = p1.next;
26             p2 = p2.next;
27         }
28         
29         return p2 == null;
30     }
31     
32     private ListNode findMiddle(ListNode head) {
33         if (head == null) {
34             return null;
35         }
36         ListNode slow = head, fast = head.next;
37         while (fast != null && fast.next != null) {
38             slow = slow.next;
39             fast = fast.next.next;
40         }
41         
42         return slow;
43     }
44     
45     private ListNode reverse(ListNode head) {
46         ListNode prev = null;
47         
48         while (head != null) {
49             ListNode temp = head.next; //为了不让head.next丢失，存tmp
50             head.next = prev;  //head指向prev，完成反转
51             prev = head;  //操作下一个节点
52             head = temp;  //操作下一个节点
53         }
54         
55         return prev;
56     }
57 }