java基于双向环形链表解决丢手帕问题的方法示例

public class Test {
  public static void main(String[] args){
      CycleLink cl=new CycleLink(5); //构造环形链表
      System.out.println("脚本之家测试结果：");
      cl.print();
      cl.setK(2); //设置从第几个小孩开始数数
      cl.setM(3); //设置数几下
      cl.play(); //开始游戏
  }
}
class Child{
  int no;
  Child nextChild;
  Child previousChild;
  public Child(int no){
    this.no=no;
  }
}
class CycleLink{
  Child first;
  Child cursor;
  int length;
  //从第几个小孩开始数
  private int k=1;
  //数几下
  private int m=1;
  //构造函数
  public CycleLink(int len){
    this.length=len;
    for(int i=1;i<=length;i++){
      Child ch=new Child(i);
      if(i==1){
        first=ch;
        cursor=ch;
      }else if(i<length){
        cursor.nextChild=ch;
        ch.previousChild=cursor;
        cursor=ch;
      }else {
        cursor.nextChild=ch;
        ch.previousChild=cursor;
        cursor=ch;
        ch.nextChild=first;
        first.previousChild=ch;
      }
    }
  }
  //打印链表
  public void print(){
    cursor=first;
    do{
      System.out.print(cursor.no+"<<");
      cursor=cursor.nextChild;
    }while(cursor!=first);
    System.out.println();
  }
  //开始游戏
  public void play(){
    Child temp;
    cursor=first;
    //先找到第k个小孩
    while(cursor.no<k){
      cursor=cursor.nextChild;
    }
    while(length>1){
      //数m下
      for(int i=1;i<m;i++){
        cursor=cursor.nextChild;
      }
      System.out.println("小孩"+cursor.no+"出局了！");
      //找到前一个小孩
      temp=cursor.previousChild;
//     temp=cursor;
//     do{
//       temp=temp.nextChild;
//     }while(temp.nextChild!=cursor);
      temp.nextChild=cursor.nextChild;
      cursor.nextChild.previousChild=temp;
      cursor=cursor.nextChild;
      length--;
    }
    System.out.println("最后一个出局的小孩是"+cursor.no);
  }
  public void setK(int k) {
    this.k = k;
  }
  public void setM(int m) {
    this.m = m;
  }
}