【LeetCode】234 - Palindrome Linked List

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? 

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 1 #include<iostream>

 2 #include<vector>

 3 using namespace std;

 4 

 5 typedef struct ListNode {

 6     int val;

 7     ListNode *next;

 8 }node;

 9  

10 bool isPalindrome1(ListNode* head);

11 bool isPalindrome2(ListNode* head);

12 ListNode* getMid(ListNode *head);

13 ListNode* reverse(ListNode *node);

14 

15 node *create(){//建立

16     node *head,*p,*s;

17     int x,cycle=1;    

18     head=new node();//head=(node*)malloc(sizeof(node));

19     p=head;

20     while(cycle){

21         cout<<"input data:";

22         cin>>x;

23         if(x!=0){

24             s=new node();

25             s->val=x;

26             p->next=s;

27             p=s;

28         }

29         else cycle=0;

30     }

31     head=head->next;

32     p->next=NULL;

33     return head;

34 }

35 int main(){

36     node *mynode=create();

37     cout<<isPalindrome2(mynode);    

38 }

39 

40 bool isPalindrome1(ListNode* head){　　//Runtime:28ms,一次遍历，取出val，装入vector再进行判断 41     if(head==NULL||head->next==NULL)return true;

42     ListNode *node=head;

43     vector<int> vec;

44     while(node!=NULL){

45         vec.push_back(node->val);

46         node=node->next;

47     }

48     for(int i=0,j=vec.size()-1;i<j;i++,j--){

49         if(vec[i]!=vec[j])return false;

50     }

51     return true;

52 };

53 

54     bool isPalindrome2(ListNode* head){　　//Runtime:28ms，拆分逆转后半链表，再同时遍历两链表 55         if(head==NULL||head->next==NULL)return true;

56         ListNode *mid=getMid(head);

57         ListNode *remid=reverse(mid);

58         while(head&&remid){

59             if(head->val!=remid->val)return false;

60             head=head->next;

61             remid=remid->next;

62         }

63         return true;

64     };

65     

66     ListNode* getMid(ListNode *head){

67         ListNode *slow=head,*fast=head;

68         ListNode *preslow;

69         while(fast->next){

70             fast=fast->next;

71             if(fast->next){

72                 fast=fast->next;

73                 preslow=slow;

74                 slow=slow->next;

75             }else{

76                 preslow=slow;

77                 slow=slow->next;

78             }    

79         }

80         preslow->next=NULL;

81         return slow;

82     }

83     ListNode* reverse(ListNode *node){

84         if(node==NULL||node->next==NULL)return node;

85         ListNode *node1=node,*node2=node1->next,*node3;    

86         while(node2){

87             node3=node2->next;

88             node2->next=node1;

89             node1=node2;

90             node2=node3;        

91         }

92         node->next=NULL;

93         return node1;

94     }