1 #include<iostream>
2 using namespace std;
3
4 /*
5 链表节点
6 */
7 struct Node
8 {
9 int value;
10 Node* next;
11 };
12 /*
13 创建一个长度为len的链表，并返回链表头结点
14 */
15 Node* init(int len)
16 {
17 Node* head = new Node[len];
18 if(head == NULL)
19 {
20 cout<<"new failed"<<endl;
21 return NULL;
22 }
23 Node* p = head;
24 for(int i = 1; i <= len ;p++, i++)
25 {
26 if( i < len )
27 {
28 p->value = i;
29 p->next = p+1;
30 }
31 else
32 {
33 p->value = i;
34 p->next = NULL;
35 }
36 }
37 return head;
38 }
39
40 /*
41 接受一个链表头结点，并将链表逆置，返回链表头结点
42 */
43 Node* reverse(Node* head)
44 {
45 if(head == NULL)
46 {
47 cout<<"链表为空"<<endl;
48 return NULL;
49 }
50 if(head->next == NULL)
51 return head;
52 Node* n_head = reverse(head->next);
53 head->next->next = head;
54 head->next = NULL;
55 return n_head;
56 }
57
58 /*
59 遍历一个链表
60 */
61 void traverse(Node* head)
62 {
63 Node* ptr = head;
64 while( ptr != NULL )
65 {
66 if(ptr->next != NULL)
67 cout<<ptr->value<<"->";
68 else
69 cout<<ptr->value<<endl;
70 ptr = ptr->next;
71 }
72 }
73 /*
74 0->1->2->3->4 ===>> 0->4->1->3->2
75 0->1->2->3 ===>> 0->3->1->2
76 思路：
77 1、找到len/2+1的节点，生成两个链表a,b
78 2、将b链表逆置
79 3、将逆置后的b一个个插入a链表中
80 */
81 void function(Node* head, int len)
82 {
83 if(head == NULL || len <= 0)
84 return;
85 int n_list_len = len/2 + 1;
86 Node* n_head = NULL;
87 Node* temp = head;
88 for(int i = 1; i < n_list_len; i++)
89 temp = temp->next;
90 n_head = temp->next;
91 temp->next = NULL;
92 Node* a = head;
93 Node* b = reverse(n_head);
94 while( b != NULL)
95 {
96 temp = b;
97 b = b->next;
98 temp->next = a->next;
99 a->next = temp;
100 a = temp->next;
101 }
102 }
103
104 int main()
105 {
106 Node* head;
107 int len = 0;
108 while(1)
109 {
110 cout<<"链表长度：";
111 cin>>len;
112 if(len <= 0)
113 {
114 cout<<"链表长度不能小于等于0"<<endl;
115 continue;
116 }
117 head = init(len);
118 traverse(head);
119 function(head,len);
120 traverse(head);
121 head = reverse(head);
122 traverse(head);
123 }
124 return 0;
125 }
