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;
}
类
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