java链表 约瑟夫问题

public class Link {


/**
* display(n,s,d); n个人，从第s个人开始，数到d出来
*/
public static void main(String[] args) {
int n = 7,d = 3, s =2;
new Link().display(n,s,d);
}



class linkNode{
int data;
linkNode next;

public linkNode(int k){
data = k;
next = null;
}
}

public  void display(int n,int s, int d){
int i=1;
linkNode head,rail,q;
head = new linkNode(i);
head.next = head;
rail=head; 
while(i<n){
i++;
q = new linkNode(i);
rail.next = q;
q.next = head;
rail = q;
}
//从s开始报数
int j=1;
linkNode p = head;
while(j<s){
p = p.next;
j++;
}//得到第s个节点
while(p != p.next){//还有1个时
j=1;
while(j < d-1){//得到第d个节点前一个节点，可删除操作
p = p.next;
j++;
}
System.out.print(p.next.data + " ");
p.next = p.next .next;//为了此操作可进行
p = p.next;
}
System.out.println(p.data);
     
}

}

2、、、、、、

public class lYSF {


/**
* 
*/
public static void main(String[] args) {

int n = 5, s=2, d =3;
YSF1 ss = new YSF1(n);
ss.disply(s,d);

}
}
class linkNode{
int data;
linkNode next;

public linkNode(int k){
data = k;
next = null;
}
}
 class YSF1{
linkNode head;
YSF1(int n){
int i = 1;
linkNode rear;
head = new linkNode(i);
rear = head;
head.next = head;
while(i<n){
i++;
linkNode p = new linkNode(i);
rear.next = p;
p.next = head;
rear = p;
}
}
public void disply(int s, int d){
int j = 0;
linkNode p = head;
while(j<s-1){
j++;
p = p.next;
}
while(p != p.next){
j=1;
while(j < d-1){
p = p.next;
j++;
}
delete(p);
p = p.next;
}
System.out.println(p.data);
}
private void delete(linkNode p) {
linkNode q = p.next;
System.out.println(q.data);
if(q == head){
head = q.next;
}
p.next = q.next;
}
}