一、引出问题
约瑟夫问题:设编号为1,2,… n的n个人围坐一圈,约定编号为k(1<=k<=n)的人从1开始报数,数到m 的那个人出列,它的下一位又从1开始报数,数到m的那个人又出列,依次类推,直到所有人出列为止,由此产生一个出队编号的序列。
例如编号为12345的环形链表,从1开始,则出队编号为:2—>4—>1—>5—>3.
二、代码实现
package Jusepfu;
public class Josepfu {
public static void main(String[] args) {
CircleSingleLinkedList circleSingleLinkedList = new CircleSingleLinkedList();
circleSingleLinkedList.addBoy(5);
circleSingleLinkedList.countBoy(1,2,5);
}
}
class CircleSingleLinkedList{
private Boy first = new Boy(-1);
public void addBoy(int nums){
if(nums < 1){
System.out.println("nums值不正确");
return;
}
Boy curBoy = null;
for(int i = 1;i <= nums ; i++){
Boy boy = new Boy(i);
if(i == 1){
first = boy;
first.setNext(first);
curBoy = first;
}else{
curBoy.setNext(boy);
boy.setNext(first);
curBoy = boy;
}
}
}
public void showBoy(){
if(first == null){
System.out.println("链表为空");
return;
}
Boy curBoy = first;
while(true){
System.out.printf("小孩的编号%d\n",curBoy.getNo());
if(curBoy.getNext() == first){
break;
}
curBoy = curBoy.getNext();
}
}
public void countBoy(int startNo,int countNum,int nums){
if(first == null || startNo < 1 || startNo > nums){
System.out.println("参数输入有误,请重新输入");
return;
}
//将helper指向环形链表的最后一个节点
Boy helper = first;
while(true){
if(helper.getNext() == first){
break;
}
helper = helper.getNext();
}
//将helper和startNo向后移到startNo处
for(int j = 0; j< startNo - 1;j++){
first= first.getNext();
helper = helper.getNext();
}
//开始出圈
while(true){
if(helper == first){
break;
}
for (int i = 0; i < countNum-1; i++) {
first = first.getNext();
helper = helper.getNext();
}
System.out.printf("小孩%d出圈\n",first.getNo());
first = first.getNext();
helper.setNext(first);
}
System.out.printf("最后留在圈中的小孩编号为%d\n",first.getNo());
}
}
class Boy{
private int no;
private Boy next;
public Boy(int no){
this.no = no;
}
public int getNo() {
return no;
}
public void setNo(int no) {
this.no = no;
}
public Boy getNext() {
return next;
}
public void setNext(Boy next) {
this.next = next;
}
}
三、测试结果
小孩2出圈
小孩4出圈
小孩1出圈
小孩5出圈
最后留在圈中的小孩编号为3