约瑟夫环问题单链表c++

可以使用单向循环链表来解决约瑟夫环问题。具体实现方法是：先构建一个有n个结点的单向循环链表，然后从头结点开始，每次数m个结点，将数到的第m个结点删除，直到链表中只剩下一个结点为止。

代码实现如下：

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

typedef struct node {
    int data;
    struct node *next;
} Node;

Node *createList(int n) {
    Node *head = NULL, *p = NULL, *tail = NULL;
    int i;
    for (i = 1; i <= n; i++) {
        p = (Node*)malloc(sizeof(Node));
        p->data = i;
        if (tail == NULL) {
            head = tail = p;
        } else {
            tail->next = p;
            tail = p;
        }
    }
    tail->next = head;
    return head;
}

void josephus(int n, int m) {
    Node *head = createList(n);
    Node *p = head, *pre = NULL;
    int i, count = 0;
    while (p->next != p) {
        count++;
        if (count == m) {
            printf("%d ", p->data);
            pre->next = p->next;
            free(p);
            p = pre->next;
            count = 0;
        } else {
            pre = p;
            p = p->next;
        }
    }
    printf("%d\n", p->data);
}

int main() {
    int n, m;
    printf("Enter n and m: ");
    scanf("%d%d", &n, &m);
    josephus(n, m);
    return 0;
}