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Josephus环类问题,java实现

2015-11-06 17:35 330 查看
  写出一个双向的循环链表,弄一个计数器,我定义的是到三的时候,自动删除当前节点,很简单。

  

package Com;

import java.util.Scanner;

/*
* 约瑟夫环问题,有n个人组成的圈,数到3的那个人出列,下个人继续从一开始
*/

public class Josephus {

public static void main(String[] args) {
Scanner s = new Scanner(System.in);
int n = Integer.parseInt(s.nextLine());
Node first = new Josephus().startRun(n );
int count = 1;
while(first.next != first) {
first = first.next;
count++;
if(count == 3) {
first.previous.next = first.next;
first.next.previous = first.previous;
first = first.next;
count = 1;
}
}
System.out.println("最后剩下来的数字为:"+first.n);
}

public Node startRun(int n) {
Node first = new Node();
first.previous = null;
first.n = n ;   //这里给链表赋值,倒叙
Node current = first;
Node last = first;
while((--n)>0) {
current.next = new Node();
current = current.next;
current.n = n;
current.previous = last;
last = current;
}
current.next = first;
first.previous = current;
return first;
}
class Node {
int n ;
Node next;
Node previous;
}
}
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