【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?

Hide Tags: Linked List Two Pointers

 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     }