双向循环链表解决约瑟夫问题

双向循环链表解决约瑟夫问题

#include <stdio.h>
#include <stdlib.h>

typedef int datatype;
typedef struct node_t
{
	datatype data;
	struct node_t * prior;
	struct node_t * next;
}link_node_t,*link_list_t;

typedef struct doublelinklist
{
	link_list_t head;
	link_list_t tail;
}double_node_t,*double_list_t;

int main(int argc, const char *argv[])
{
	int i;
	int all_num = 8;//猴子总数
	int start_num = 3;//从3号猴子开始数
	int kill_num = 3;//数到几杀死猴子 
	link_list_t h = NULL;
	link_list_t pdel = NULL;//用来指向被杀死猴子的节点
	printf("请您输入猴子的总数，开始号码，出局号码:\n");
	scanf("%d%d%d",&all_num,&start_num,&kill_num);
	//1.创建一个双向的循环链表
	double_list_t p = (double_list_t)malloc(sizeof(double_node_t));//申请头指针和尾指针
	if(NULL == p)
	{
		perror("malloc failed");
		return -1;
	}
	p->head = p->tail = (link_list_t)malloc(sizeof(link_node_t));
	if(NULL == p->tail)
	{
		perror("p->tail malloc failed");
		return -1;
	}
	p->head->data = 1;
	p->head->prior = NULL;
	p->head->next = NULL;
	//将创建n个新的节点，链接到链表的尾
	for(i = 2; i <= all_num; i++)
	{
		link_list_t pnew = (link_list_t)malloc(sizeof(link_node_t));
		if(NULL == pnew)
		{
			perror("pnew malloc failed");
			return -1;
		}
		pnew->data = i;
		pnew->prior = NULL;
		pnew->next = NULL;
		//(1)将新的节点链接到链表的尾
		p->tail->next = pnew;
		pnew->prior = p->tail;
		//(2)尾指针向后移动，指向当前链表的尾
		p->tail = pnew;
	}
	//(3)形成双向循环链表 
	p->tail->next = p->head;
	p->head->prior = p->tail;
	//调试程序 
#if 0
	while(1)
	{
		printf("%d\n",p->head->data);
		p->head = p->head->next;
		sleep(1);
	}
#endif
	//2.循环进行杀死猴子
	h = p->head;
	//(1)先将h移动到start_num处，也就是开始数数的猴子号码处
	for(i = 0; i < start_num-1; i++)
		h = h->next;
	while(h->next != h)//当h->next == h 就剩一个节点了，循环结束
	{
		//(2)将h移动到即将杀死猴子号码的位置
		for(i = 0; i < kill_num-1; i++)
			h = h->next;
		//(3)进行杀死猴子，经过上面的循环后，此时的h指向即将杀死的猴子
		h->prior->next = h->next;
		h->next->prior = h->prior;
		pdel = h;//pdel指向被杀死猴子的位置
		printf("kill is -------%d\n",pdel->data);
		h = h->next;//需要移动，从杀死猴子后的下一个位置开始数
		free(pdel);
	}
	printf("猴王是%d\n",h->data);
	return 0;
}