【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 };