用链表写约瑟夫环

约瑟夫环自己以前写过，现在想用数据结构实现一次。

先贴上自己先前用数组实现的代码：

/*约瑟夫环有n个人(n<=1000)，用1,2,...,n编号，顺序排列，并首尾相连围成一圈。从第一个人开始报数(从1到4)，凡报到4的人退出圈子，且后面的人继续报数（同样从1到4报数），问最后留下的是原来第几号的那一位输入人数输出最后剩下的人。*/#include<stdio.h>int main(){int a[1001],i,j,k,n;scanf("%d",&n);for(i=1; i<=n; i++)a[i]=i;for(i=1,j=1,k=n; ;j++,i++)    {         if(k==1)break;         if(i>n)i-=n;         if(a[i]==0) {j--;continue; }if(j%4==0){if(a[i]!=0){a[i]=0;k--;}else{j--;continue;}}}for(i=1; i<=n; i++){if(a[i]!=0){printf("%d",a[i]);break;}}return 0;}

下面是链表实现的约瑟夫环

/*53人从1-4开始报数，。。。（和前面雷同）*/#include<stdio.h>#include<stdlib.h>typedef struct node{int data;int visited;struct node* next;}*linklist,list;int length = 53;linklist creat(){int number;linklist temp,p;linklist head;head = (linklist)malloc(sizeof(list));head->data =1;head->visited = 0;head->next = NULL;temp = head;for(number=2;number<=53;number++){p = (linklist)malloc(sizeof(list));p->data = number;p->visited = 0;p->next = temp->next;temp->next = p;temp = temp->next;}temp->next = head;return head;}int work(linklist plist){linklist head = plist;linklist tmp = plist->next;int i=4;int flag=1;while(length != 1){if(head->visited){head = head->next;continue;}if(flag == i && head->visited == 0){length--;head->visited = 1;head = head->next;flag=1;}if(head->visited){head = head->next;continue;}head = head->next;flag++;}if(plist->visited == 0){printf("%d\n",plist->data);return 1;}while(tmp->data != plist->data){tmp = tmp->next;if(tmp->visited == 0){printf("%d\n",tmp->data);break;}}return 1;}int main(){linklist head =NULL;head = creat();work(head);return 0;}