设单链表中存放着n个字符，每个节点保存一个字符。试编写算法，判断该字符串是否有中心对称关系。

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdlib>
 4 using namespace std;
 5 struct node* create1(string);
 6 struct node* create2(string);
 7 struct node
 8 {
 9     char elem;
10     struct node* next;
11 };
12 int main()
13 {
14     char str[40];
15     cin >> str;
16     struct node* p1 = create1(str);
17     struct node* p2 = create2(str);
18     //while(strcmp(p1->elem,p2->elem))
19     for (int i = 0; i < strlen(str); i++)
20     {
21         if (p1->elem == p2->elem)
22         {
23             p1 = p1->next;
24             p2 = p2->next;
25         }
26         else 
27         {
28             cout << str << " is not symmetrical";
29             break;
30         }
31     }
32     if (p1 == NULL && p2 == NULL)
33         cout << str << " is symmetrical";
34     return 0;
35 }
36 struct node* create1(string s)
37 {
38     int n = s.length();
39     struct node* head = NULL, * p1, * p2;
40     p2 = p1 = (struct node*)malloc(sizeof(struct node));
41     for (int i = 0; i < n; i++)
42     {
43         if (head == NULL)
44             head=p1;
45         else
46             p2->next = p1;
47         p1->elem = s[i];
48         p2 = p1;
49         //p1->elem = s[i];
50         p1 = (struct node*)malloc(sizeof(struct node));
51     }
52     p2->next = NULL;
53     free(p1);
54     return head;
55 
56 }
57 struct node* create2(string s)
58 {
59     int n = s.length();
60     struct node* head = NULL, * p1;
61     p1= (struct node*)malloc(sizeof(struct node));
62     for (int i = 0; i < n; i++)
63     {
64         p1->elem = s[i];
65         p1->next = head;
66         head = p1;
67         p1 = (struct node*)malloc(sizeof(struct node));
68     }
69     free(p1);
70     return head;
71 }

标签：node,p2,单链,struct,中心对称,elem,next,p1,表中
来源： https://www.cnblogs.com/xu-chen-313/p/14598870.html