循环链表解决约瑟夫环问题

约瑟夫环问题，已知n个人，从编号为k的人开始报数，数到m的那个人出列；他的下一个又从1开始报数， 数到m的那个人又出列，依此重复，直到所有的人出列。思路如下：
(1)建立一个具有n个结点、无头结点的循环链表。
(2)确定第一个报数人的位置

(3)不断地从链表中删除结点，直到链表为空。

typedef struct node {int num;struct node *next;}Node,*PNode;void JOSEPHUS(int n, int k, int m){//建立没有头结点的循环链表PNode curr = NULL;PNode pHead = (PNode)malloc(sizeof(Node));if (pHead == NULL){printf("内存分配失败！\n");exit(EXIT_FAILURE);}pHead->num = 1;pHead->next = pHead;curr = pHead;for (int i = 2;i <= n;++i) {PNode pNew = (PNode)malloc(sizeof(Node));pNew->num = i;curr->next = pNew;curr = pNew;pNew->next = pHead;}//把当前指针移动到第一个报数的人curr = pHead;while (curr->num != k)curr = curr->next;while (curr->next != curr){for (int i = 1;i < m;++i) {curr = pHead;pHead = pHead->next;}printf("%d ",pHead->num);curr->next = pHead->next;free(pHead);pHead = curr->next;}printf("%d \n",pHead->num);free(pHead);}