线性表练习之Josephus

题目描述:

利用循环链表实现约瑟夫问题。约瑟夫问题如下:已知n个人(n>=1)围坐一圆桌周围,从1开始顺序编号。从序号为1的人开始报数,顺时针数到m的那个人出列;他的下一个人又从1开始报数,数到m的那个人又出列;依此规则重复下去,直到所有人全部出列。请问最后一个出列的人的编号。

输入描述:
一行,人数n和整数m,中间用空格隔开

输出描述:
一行,最后一个出列的人的编号

样例输入:
5 3
样例输出:
4

全代码:

#include <iostream>
using namespace std;

class Node
{
public:
 Node(int i) :data(i), next(0) {}
 int data;
 Node *next;
};

Node * CLinkList(int n)
{
 Node *rear = new Node(1);
 Node *r = rear;
 for (int i = 2; i <= n; i++)
 {
   Node * s = new Node(i);
   r->next = s;// 生成新结点
   r = r->next;
 }
 r->next = rear;
 return rear;
}

void Output(Node *rear, int m)
{
 Node *r = rear;
 Node *s = r->next;
 int i = 1;
 while (s != r)
 {
  if (i != m)
  {
   r = s;
   s = r->next;
   i++;
  }
  else
  {
   r->data = s->data;
   r->next = s->next;
   delete s;
   s = r->next;
   i = 1;
  }
 }
 cout << r->data << endl;
}

int main()
{
 cout << "设定参与人数和循环数分别为:";
 int m, n;
 cin >> n >> m;
 Node *first = CLinkList(n);
 Output(first, m);
 return 0;
}

class Node
{
public:
 Node(int i) :data(i), next(0) {}
 int data;
 Node *next;
};

浅析:
(1)data 的值为i

单循环链表的建立

Node * CLinkList(int n)
{
 Node *rear = new Node(1);
 Node *r = rear;
 for (int i = 2; i <= n; i++)
 {
   Node * s = new Node(i);
   r->next = s;// 生成新结点
   r = r->next;
 }
 r->next = rear;
 return rear;
}

浅析:
循环链表rear的重要性

删除和输出

void Output(Node *rear, int m)
{
 Node *r = rear;
 Node *s = r->next;
 int i = 1;
 while (s != r)
 {
  if (i != m)
  {
   r = s;
   s = r->next;
   i++;
  }
  else
  {
   r->data = s->data;
   r->next = s->next;
   delete s;
   s = r->next;
   i = 1;
  }
 }
 cout << r->data << endl;
}

浅析:
计数,删除,循环,直到只剩一人

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转载自blog.csdn.net/FirstOfSky/article/details/105275343