Data Structure—Joseph Ring Problem (C Language Version)

Table of contents

First what is a joseph ring

Joseph ring implementation

1. Create structure variables

2. Initialize the linked list

3. Construct a circular linked list

4. Delete linked list

5. Complete code and comments explanation

First what is a joseph ring

Joseph's ring is a classic problem in circular linked lists; title description: n people form a circle, start counting from the first person, people who count to m get out of the queue, and then the next person starts counting from 1 again . The person who counts to m will go out of the circle, and so on, until everyone is out of the circle;

Assume that 10 people form a circle, numbered from 1 to 10, and the number is reported in ascending order. The person who reports 3 is out. The process diagram is as follows

Joseph ring implementation

I personally prefer circular linked lists;

1. Create structure variables

typedef struct Node{
    int data;  //数据域
    struct Node* next;  //指针域
}Node;

2. Initialize the linked list

Node* Create(){
	 Node* head;
     head = (Node*)malloc(sizeof(Node));
     if (head == NULL) {
         exit(1);
     }
     head->next = NULL;
     return head;
}

3. Construct a circular linked list

Create a temporary node tail , assign the head node head to tail , insert a new node p , let the next of tail point to p , and the next of p point to the next node of head , which is node p ; at the same time, let tail move to p position to prepare for the insertion of the next node; in this way, a circular linked list is initially completed;

code block

Node* Push(Node* head,Node* tail,int 1){
    p->data=i;
    tail->next =p;
    p->next = head->next;
    tail = p;
    return head;
}

The most important thing is to move the tail to the new node position every time you insert a new node!

4. Delete linked list

void Print(Node* head){
    Node* q=head;
    q->next = p->next;
    printf("%d ", p->data);
    free(p);
    p = q->next;
}

5. Complete code and comments explanation

#include<stdio.h>
#include<stdlib.h>
typedef struct Node{
    int data;
    struct Node* next;
}Node;
//创建结构体变量
Node* Create(){
	 Node* head;
     head = (Node*)malloc(sizeof(Node));
     if (head == NULL) {
         exit(1);
     }
     head->next = NULL;
     return head;
}
int main()
{
     int n, m,i;
     Node * tail, * p, * q;
     Node* head=Create();
     scanf("%d %d", &n, &m);  //输入n个人围一圈及报数m的人出局
//判断如果插入的数据为0或者以报数0为出局，则结束操作
     if (n == 0 || m == 0) {
         return 0;
     }
     else {
         tail = head;  //开始没有数据，故尾结点tail与头结点重合
         for ( i = 0; i < n; i++) {
             p = (Node*)malloc(sizeof(Node));  //插入新结点需，先申请动态内存
             if (p == NULL) {        //判断动态内存是否申请成功
                 printf("申请失败!");
                 exit(1);
             }
             p->data = i+1; //以下4步为插入新结点及数据的操作，具体分析请看上面构建循环链表
             tail->next =p;
             p->next = head->next;
             tail = p;
         }
     }
     p =head->next;  //插入完数据后，将最后一个结点的临时结点移到第一个数据处
     q =tail;   //然后临时结点到尾结点处
     i = 1;
     while (p != q) {      //首尾结点是否重合，重合则表示只剩一个数据，结束循环
         if (i == m) {     //对报数m的人进行出局操作
             q->next = p->next;  //以下四步为删除操作
             printf("%d ", p->data);
             free(p);     //一定记得将删除链表处的内存释放，以免内存内存泄漏
             p = q->next;
             i = 1;     //删除后，重新从1开始报数
         }
         else {   //没有报数到m，则p，q结点都往后移一位
             q = p;  //先q移到p的位置
             p = q->next;  //然后p移到q的下一个位置
             i++;
         }
     }
     printf("%d", p->data);  //打印最后一位出局的人的号数
     return 0;
}