华为面试题——约瑟夫问题的C++简单实现（循环链表）

/*

author:jiangxin

Blog:http://blog.csdn.net/jiangxinnju

Function:method of Josephus question

*/

#include <iostream>

using namespace std;

struct node

{

int seq;

node *next;

};

typedef struct node NODE;

void test_Josephus()

{

/*如果共同拥有n人。从第s个人開始数数，每数到m该人出列，后面的人又一次開始数。知道所有人出列*/

int n,s,m;

NODE *head,*last,*current,*prev;

cout << "Input the n,s,m(separate with space):";

cin >> n >> s >> m;

for(int i=1;i<=n;i++) //建立循环链表

{

current = new NODE;

current->seq = i;

current->next = head;

if(i == 1)

{

head = current;

last = current;

}

else

{

last->next = current;

last = last->next;

}

}

current = head; //遍历循环链表，输出序列

do

{

cout << current->seq << " ";

current = current->next;

}while(current!=head);

current = head; //将current置于第s个位置

for(int i=1;i<s;i++)

{

current = current->next;

}

cout << endl;

for(int i=1;i<=n;i++) //共循环n轮，得到一个总体的输出序列

{

for(int j=1;j<m;j++)

{

prev = current;

current = current->next;

}

cout << current->seq << " ";

prev->next = current->next;

delete current;

current = prev->next;

}

}