魔术师发牌问题简单实现-循环链表

#include<stdio.h>

#include<stdlib.h>

#define CardNumber 13

typedef struct node

{

int data;

struct node *next;

}sqlist,*linklist;

//创建13个节点的循环链表

linklist createLinkList()

{

linklist head=NULL;

linklist s,r;

r=head;

for(int i=1;i<=CardNumber;i++)

{

s=(linklist)malloc(sizeof(sqlist));

s->data=0;

if(head==NULL)

head=s;

else

r->next=s;

r=s;

}

r->next=head;

return head;

}

//发牌顺序

void Magician(linklist head)

{

linklist p;

int countnumber=2;

p=head;

p->data=1;

while(1)

{

for(int j=0;j<countnumber;j++)

{

p=p->next;

if(p->data!=0)

{

p=p->next;

j--;

}

}

if(p->data==0)

{

p->data=countnumber;

countnumber++;

if(countnumber==14)

break;

}

}

}

void DestoryList(linklist *list)

{

linklist p=*list;

linklist q=NULL;

int i=1;

while(i<=CardNumber)

{

q=p;

p=p->next;

free(q);

i++;

}

}

int main()

{

linklist p;

p=createLinkList();

Magician(p);

printf("按下列顺序排放:\n");

for(int i=0;i<CardNumber;i++)

{

printf("黑桃%d",p->data);

p=p->next;

}

DestoryList(&p);

return 0;

}