循环链表的实现
单链表只有向后结点，当单链表的尾链表不指向NULL，而是指向头结点时候，形成了一个环，成为单循环链表，简称循环链表。当它是空表，向后结点就只想了自己，这也是它与单链表的主要差异，判断node->next是否等于head。
代码实现分为四部分：
- 初始化
- 插入
- 删除
- 定位寻找
代码实现：
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
|
void ListInit(Node *pNode){ int item; Node *temp,*target; cout<<"输入0完成初始化"<<endl; while(1){ cin>>item; if(!item) return ; if(!(pNode)){ //当空表的时候，head==NULL pNode = new Node ; if(!(pNode)) exit(0);//未成功申请 pNode->data = item; pNode->next = pNode; } else{ // for(target = pNode;target->next!=pNode;target = target->next) ; temp = new Node; if(!(temp)) exit(0); temp->data = item; temp->next = pNode; target->next = temp; } }} void ListInsert(Node *pNode,int i){ //参数是首节点和插入位置 Node *temp; Node *target; int item; cout<<"输入您要插入的值:"<<endl; cin>>item; if(i==1){ temp = new Node; if(!temp) exit(0); temp->data = item; for(target=pNode;target->next != pNode;target = target->next) ; temp->next = pNode; target->next = temp; pNode = temp; } else{ target = pNode; for (int j=1;j<i-1;++j) target = target->next; temp = new Node; if(!temp) exit(0); temp->data = item; temp->next = target->next; target->next = temp; }}void ListDelete(Node *pNode,int i){ Node *target,*temp; if(i==1){ for(target=pNode;target->next!=pNode;target=target->next) ; temp = pNode;//保存一下要删除的首节点 ,一会便于释放 pNode = pNode->next; target->next = pNode; delete temp; } else{ target = pNode; for(int j=1;j<i-1;++j) target = target->next; temp = target->next;//要释放的node target->next = target->next->next; delete temp; }}int ListSearch(Node *pNode,int elem){ //查询并返回结点所在的位置 Node *target; int i=1; for(target = pNode;target->data!=elem && target->next!= pNode;++i) target = target->next; if(target->next == pNode && target->data!=elem) return 0; else return i;} |
约瑟夫问题
约瑟夫环（约瑟夫问题）是一个数学的应用问题：已知n个人（以编号1，2，3...n分别表示）围坐在一张圆桌周围。从编号为k的人开始报数，数到m的那个人出列；他的下一个人又从1开始报数，数到m的那个人又出列；依此规律重复下去，直到圆桌周围的人全部出列。这类问题用循环列表的思想刚好能解决。
注意：编写代码的时候，注意报数为m = 1的时候特殊情况
|
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
|
#include<iostream>#include<cstdio>using namespace std;typedef struct Node{ int data; Node *next;};Node *Create(int n){ Node *p = NULL, *head; head = new Node; if (!head) exit(0); p = head; // p是当前指针 int item=1; if(n){ int i=1; Node *temp; while(i<=n){ temp = new Node; if(!temp) exit(0); temp->data = i++; p->next = temp; p = temp; } p->next = head->next; } delete head; return p->next;}void Joseph(int n,int m){ //n为总人数，m为数到第m个的退出 m = n%m; Node *start = Create(n); if(m){//如果取余数后的m!=0，说明 m!=1 while(start->next!=start){ Node *temp = new Node; if(!temp) exit(0); for(int i=0;i<m-1;i++) // m = 3%2 = 1 start = start->next; temp = start->next; start->next = start->next->next; start = start->next; cout<<temp->data<<" "; delete temp; } } else{ for(int i=0;i<n-1;i++){ Node *temp = new Node; if(!temp) exit(0); cout<<start->data<<" "; temp = start; start = start->next; delete temp; } } cout<<endl; cout<<"The last person is:"<<start->data<<endl;}int main(){ Joseph(3,1); Joseph(3,2); return 0;} |
感谢阅读，希望能帮助到大家，谢谢大家对本站的支持！
原文链接：http://www.cnblogs.com/AsuraDong/p/6986930.html
