【LeetCode 234】Palindrome Linked List

Given a singly linked list, determine if it is a palindrome.

思路：

　　用快慢指针找到链表中点，反转后半部分链表，然后与前半部分进行匹配，随后将链表恢复原状（本题没有这个要求，具体情况具体对待）。

C++：

 1 /**
 2  * Definition for singly-linked list.
 3  * struct ListNode {
 4  *     int val;
 5  *     ListNode *next;
 6  *     ListNode(int x) : val(x), next(NULL) {}
 7  * };
 8  */
 9 class Solution {
10 public:
11
12     ListNode* reverseList(ListNode *listHead)
13     {
14         ListNode *preNode = NULL;
15         ListNode *curNode = listHead;
16         ListNode *reverseHead = NULL;
17
18         while(curNode != NULL)
19         {
20             ListNode *nextNode = curNode->next;
21
22             if(nextNode == NULL)
23             {
24                 reverseHead = curNode;
25             }
26
27             curNode->next = preNode;
28             preNode = curNode;
29             curNode = nextNode;
30         }
31
32         return reverseHead;
33     }
34
35     bool isPalindrome(ListNode* head) {
36
37         if(head == 0 || head->next == 0)
38             return true;
39
40         ListNode *p1 = head;
41         ListNode *p2 = head;
42         while(p2->next != 0 && p2->next->next != 0)
43         {
44             p1 = p1->next;
45             p2 = p2->next->next;
46         }
47
48         ListNode *rhead = reverseList(p1->next);
49         p1->next = 0;
50
51         p1 = head;
52         p2 = rhead;
53         while(p2 != 0)
54         {
55             if(p2->val != p1->val)
56                 return false;
57
58             p1 = p1->next;
59             p2 = p2->next;
60         }
61
62         p1 = head;
63         while(p1->next != 0)
64             p1 = p1->next;
65
66         ListNode *mhead = reverseList(rhead);
67         p1->next = mhead;
68
69         return true;
70     }
71 };