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