java-13个坏人和13个好人站成一圈，数到7就从圈里面踢出一个来，要求把所有坏人都给踢出来，所有好人都留在圈里。请找出初始时坏人站的位置。
import java.util.ArrayList;import java.util.List;public class KickOutBadGuys {/** * 题目：13个坏人和13个好人站成一圈，数到7就从圈里面踢出一个来，要求把所有坏人都给踢出来，所有好人都留在圈里。请找出初始时坏人站的位置。 * Maybe you can find out the mathematical rule behind the question. * But we try to figure it out in Java. * It's easy to have Circular Linked List in mind. * Of course we can use Java's 'LinkedList'. * But I implement my "Circular Linked List" for practice. */private static final int MAX=26;private static final int STEP=7;private static final int NUM=13;//number of bad guyspublic static void main(String[] args) {int[] guys=new int[MAX];for(int i=0;i<MAX;i++){guys[i]=i;}Node head=initialCircularList(guys);printCircularList(head);List<Node> badGuys=new ArrayList<Node>();head=kickOutBadGuys(head,badGuys);printCircularList(head);//print location of bad guysSystem.out.print("locations of bad guys are:");for(Node node:badGuys){System.out.print(node.id+" ");}}public static Node kickOutBadGuys(Node head,List<Node> list){Node node=head;Node previous=node;int i=0;while(i<NUM){int step=STEP;while(step>0){previous=node;node=node.next;step--;}list.add(node);previous.next=node.next;i++;}return previous;//No matter what 'previous' is,we can traverse the 'CircularList',starting from 'previous'}public static Node initialCircularList(int[] data){if(data==null||data.length==0){return null;}int len=data.length;Node tail=new Node(data[len-1]);Node head=tail;int i=len-2;while(i>=0){Node p=new Node(data[i]);p.next=head;//尾插法tail.next=p;head=p;i--;}return head;}public static void printCircularList(Node head){if(head==null){return;}System.out.print(head.id+" ");Node node=head.next;while(node!=null&&node!=head){//end condition is not "node.next==null" but "node==head"System.out.print(node.id+" ");node=node.next;}System.out.println();}private static class Node{int id;Node next;Node(int id){this.id=id;}}}