用链表实现约瑟夫环

约瑟夫环问题，这是一个很经典算法，处理的关键是：伪链表

问题描述：N个人围成一圈，从第一个人开始报数，报到m的人出圈，剩下的人继续从1开始报数，报到m的人出圈；如此往复，直到所有人出圈。（模拟此过程，输出出圈的人的序号）

在数据结构与算法书上，这个是用链表解决的。 https://blog.youkuaiyun.com/weixin_38214171/article/details/80352921

 

 

 

 

#include <stdio.h>
#include <stdlib.h>
//#define N 10
//#define M 3
typedef struct node{
	int n;
	struct node *next;
}S_T;

S_T *mk_node(int n, S_T *next)
{
	S_T *p = malloc(sizeof(S_T));
	if(p != NULL){
		p->n = n;
		p->next = next;
	}
	return p;
}

int main(void)
{
	S_T *head, *del;
	int i;
	int N, M;
	scanf("%d %d", &N, &M);

	head = mk_node(1, NULL);
	head->next = head;
	
	for(i = 2; i <= N; i++)
    {
		//head = head->next = creat_node(i,head->next);
		head->next = creat_node(i,head->next);
		head = head->next;
	}

	while(head != head->next)
    {
		for(i = 1; i < M; i++)
        {
			head = head->next;
		}
		del = head->next;
		head->next = head->next->next;
		printf("%d out\n", del->n);
		free(del);
	}
	printf("%d survie\n", head->n);
	return 0;
}
/*
akaedu@akaedu-G41MT-D3:~/lin/lb$ ./myysfh 
10 3
3 out
6 out
9 out
2 out
7 out
1 out
8 out
5 out
10 out
4 survie*/

 

 

 

/*
N个人围成一圈，从第一个人开始报数，报到m的人出圈，剩下的人继续从1开始报数，报到m的人出圈；如此往复，直到所有人出圈。
*/


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


//#define N 20
//#define M 5

typedef struct link_node
{
	int n;
//	char name[64]={0};
	struct link_node *next; 
}LinkNode;

//LinkNode *creat_node(int num, char *strname, LinkNode *next)
LinkNode *creat_node(int num, LinkNode *next)
{
	LinkNode *p = malloc(sizeof(LinkNode));

	if(p != NULL)
	{
		p->n = num;
		//strcpy(p->name,strname);
		p->next = next;		
	}

	return p;
}




int main(void)
{
	int i=0;
	int N=0;
	int M=0;
	LinkNode *head, *del;


	
	scanf("%d %d", &N, &M);

	head = creat_node(1, NULL);
	head->next = head;

	for(i=2; i <= N; i++)
	{
		//head = head->next = creat_node(i,head->next);
		head->next = creat_node(i,head->next);
		head = head->next;
	}
	
	while(head != head->next)
	{
		for(i = 1; i < M; i++)
		{
			head = head->next;
		}
		del = head->next;
		head->next = head->next->next;
		printf("%d out\n", del->n);
		free(del);
	}
	printf("%d survie\n", head->n);
	return 0;

}



/*
akaedu@akaedu-G41MT-D3:~/lin/lb$ ./myysfh 
10 3
3 out
6 out
9 out
2 out
7 out
1 out
8 out
5 out
10 out
4 survie*/