public ListNode deleteDuplicates(ListNode head) {
if(head==null||head.next==null)
throw new IllegalArgumentException("the number of nodes must be more than one.");
ListNode head0=head;
ListNode cur=head0.next ;
if(head0==cur){
head0=cur;
}
while(cur.next!=null){
System.out.println(cur);
if(cur==cur.next){
ListNode node=cur.next;
cur.next=node.next;
node.next=null;
}else
cur=cur.next;
}
return head0;
}
方法运行链表不变化 啊

给你个提示，90%的链表题，加一个虚头节点会好做很多：
ListNode vhead = new ListNode(0);
vhead.next=head;
var prev=vhead;
var p=head;