josephus 约瑟夫的循环链表实现

最新推荐文章于 2020-10-10 14:52:37 发布

原创最新推荐文章于 2020-10-10 14:52:37 发布 · 1.4k 阅读

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#iostream #struct #null #delete #access

本文介绍了一个经典的约瑟夫环问题，并通过C++代码详细展示了如何创建一个单链表来模拟这一问题的过程，最终找到最后一个留在圈中的人。

//有n个人围坐在一个圆桌周围，现从第s个人开始报数，数到第m
//的人出列，然后从出列的下一个人重新开始报数，数到第m的人
//又出列，…，如此反复直到所有的人全部出列为止。

#include <iostream>
#include <iomanip>
using namespace std;

typedef struct LNode{
int NO;
struct LNode * next;
}Lnode, *Josephus;

Josephus InitJosephus()
{
LNode * head = new LNode;
head->next = head;
head->NO = 0;
return head;
}

Josephus CreateJosephus(int enjoyIn)
{
LNode * node = NULL;
LNode * head = InitJosephus();
LNode * forword = head;

for( int i = 1; i <= enjoyIn; i++ )
{
  node = new LNode;
  node->NO = i;
        forword->next = node;
  forword = forword->next;
  node->next = NULL;
}
forword->next = head->next;
    return head;
}

void JoseProcess( Josephus &Jose, int position, int cycle)
{
Josephus p = Jose->next;
Josephus qTemp = NULL;
int temp;
for(int i = 0; i < position-1; i++ )
{
p = p->next;
}
cout<<"Out of the queue... "<<endl;

if(cycle == 1 && p->next != p)
{
    qTemp = p;
  while( qTemp->next != p )
  qTemp = qTemp->next;
}
if(cycle == 1 && p->next == p)
{
  cout<<"The winner is "<< p->NO <<" child! "<<endl;

}
while( p->next != p )
{

  for(int j = 1; j < cycle; j++ )
  {
   qTemp = p;
   p = p->next;
  }

temp = p->NO;

        qTemp->next = p->next;
  delete p;
  p = qTemp->next;
  cout<<temp<<endl;
}
cout<<"The winner is "<< p->NO <<" child! "<<endl;
}

void main()
{
int enjoyIn;
int position;
int cycle;

cout<<"the total of the children : ";
cin>>enjoyIn;
cout<<endl<<"the position you want to access : ";
cin>>position;
cout<<endl<<"the cycle you want : ";
cin>>cycle;
cout<<endl;

Josephus Jose = CreateJosephus( enjoyIn );
JoseProcess( Jose, position, cycle);
}