约瑟夫环问题

问题描述：假设有N个小孩按照序号1，2，，，N围坐成一圈，从第一个小孩开始报数，每次报到n的人退出，接着从下一个人重新开始从1开始报数，下一次再报到n的人退出，求最后一个留下的人；

数组实现：

public class huan {

    /**
     * @param args
     */
    public static void main(String[] args) {
        String[] s = new String[]{"0a","1b","2c","3d","4e","5f","6g","7h","8i","9j","10k","11l","12m","13n","14o","15p"};
        int k = 3;
        while(s.length != 0){
            int i = 1;
            while(i <= 2){
                if(++k > s.length-1) {//循环n次的时候，每次递增时进行下表越界判断   或者k=k%(s.length)
                    k = 0;
                }
                if(s[k] != null) {
                    i++;
                }
            }
            System.out.print(s[k] + ' ');
            s[k] = null;
        }
    }
}

输出：5f 7h 9j 11l 13n 15p 1b 3d 6g 10k 14o 2c 8i 0a 12m 4e 

循环链表实现：

package com.iteye.ljmdbc7a;

import java.util.Scanner;

/**
 * 循环列表的Java实现，解决约瑟夫环问题
 * @author LIU
 *
 */
public class LinkedList
{
    //定义结点，必须是static
    static class Node
    {
        int data;
        Node next;
        Node(int arg1)
        {
            this.data = arg1;
        }
    }
    public static void main(String[] args)
    {
        int n = 0,m = 0;//定义总人数n，和出圈数字m
        //输入n和m
        System.out.println("输入总人数n，出圈数字m");
        Scanner scanner = new Scanner(System.in);
        n = scanner.nextInt();
        m = scanner.nextInt();
        
        //初始化循环列表,头结点first和尾结点p
        Node first = new Node(1);
        first.next = first;
        Node p = first;
        for(int i=2; i<=n; i++)
        {
            Node temp = new Node(i);
            temp.next = p;
            p.next = temp;
            p = p.next;
        }
        p.next = first;//尾接头形成循环链表（p为尾结点）
        
        //执行出圈操作
        System.out.println("出圈顺序为:");
        while(p != p.next)
        {
            //下面for循环后，p是第m个结点的前一个结点
            for(int i=1; i<m; i++)
                p = p.next;
            //删除第m个结点
            System.out.print(p.next.data+" ");
            p.next = p.next.next;
        }
        System.out.print("
幸运者是:"+p.data);
    }
    
}