用java实现链表并解决约瑟夫环问题

package 栈和队列;/** *  * 数到3的人出列 ，看最后能剩下谁 * @author wangmeng * */class Node {public  Object data;public  Node next;public Object getData() {return data;}public void setData(Object data) {this.data = data;}public Node getNext() {return next;}public void setNext(Node next) {this.next = next;}public Node() {super();// TODO Auto-generated constructor stub}public Node(Object data, Node next) {super();this.data = data;this.next = next;}}public class 用java实现链表并解决约瑟夫环问题 {  public static void main(String[] args) {  Node node1 = new Node("1",null);   Node node2 = new Node("2",null);   Node node3 = new Node("3",null);   Node node4 = new Node("4",null);   Node node5 = new Node("5",null);   Node node6 = new Node("6",null);   Node node7 = new Node("7",null);         Node header = node1;   node1.next = node2;   node2.next = node3;   node3.next = node4;   node4.next = node5;   node5.next = node6;   node6.next = node7;   node7.next = node1;      检测(header);}private static void 检测(Node header) {Node n = header;Node lastOne ;//表示上一个节点int  count  = 1 ;System.out.print("依次出列的顺序为：");while(n != n.next){//最后环中只剩下一个人，那肯定是它的下一个就是它本身  即  n=n.nextlastOne = n;n = n.next;count ++ ;if(count == 3){//若数到3，出列lastOne.next = n.next;//上一个的节点的下一个就是当前的下一个   即lastOne.next = n.next;count = 0;//count清 0 System.out.print(n.data   + "   ");//打印出列的元素}}System.out.println();System.out.println("最后剩下的是:" + n.data);}             }运行结果：依次出列的顺序为：3   6   2   7   5   1   最后剩下的是:4