Remove Duplicates from Sorted List II

Q:
Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.
For example,
Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

A:
-------------------------4th pass --------用递归的思想做的-------

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head == null || head.next == null)
            return head;
        if(head.val == head.next.val){
            int findVal = head.val;
            while(head !=null && head.val == findVal)
                head=head.next;
            return deleteDuplicates(head);
        }else{
            head.next = deleteDuplicates(head.next);
            return head;
        }
    }
}

--------------------3 rd Pass ------------------ 一遍AC  ，不错----------------------

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode header = new ListNode(0);
        header.next = head;
        ListNode preNode = header; // pre node 
        ListNode curNode = head; // current node
        while(curNode!= null && curNode.next != null){
            if(curNode.val == curNode.next.val){
                // delete all values that 
                int value = curNode.val;
                while(preNode.next != null && preNode.next.val == value){
                    preNode.next = preNode.next.next;
                }
                curNode=preNode.next;
            }else{
                preNode = preNode.next;
                curNode = curNode.next;
            }
        }
        return header.next;
    }
}

-------------------------------第二遍---------------------------
思路就是，各自用两个header， 旧的每次取下第一个，记录下value，查看下面的是否相同，并设置一个boolean 标记。  然后add 到新list 的tail上去。

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        // build another list, and move only those with no ducplicats
        ListNode header1 = new ListNode(0);
        header1.next = head;
        
        ListNode header = new ListNode(0);
        ListNode tail =header;
        
        int oldVal;
        boolean findDuplicate;
        while(header1.next!= null){
            if( header1.next.next == null ){ // if only one node left
                tail.next = header1.next;
                tail =tail.next;
                break;
            }
            oldVal = header1.next.val;
            findDuplicate = false; 
            ListNode tmpNode = header1.next;
            header1.next = header1.next.next;  //delete the first node
            while(header1.next != null && header1.next.val == oldVal){
                //delete first node in header1 list
                findDuplicate = true;
                header1.next = header1.next.next;
            }
            
            if( ! findDuplicate){
                tail.next = tmpNode;
                tail = tail.next;
            }
            
            
        }
        // set tail.next as null
        tail.next = null;
        return header.next;
    }
}

--------------------1st  --------每次detach head, 然后加到结果的tail 上。---------------- -----------------

public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode header = new ListNode(0),tail = header;
        while(head != null){
            // delete all the duplicate node
            if(head.next ==null || head.val != head.next.val){
                //detach the head
                tail.next = head;
                tail = tail.next;
                head = head.next;
                tail.next = null;
            }else{ // delete all nodes with the same value of node
                int duplicateVal = head.val;
                while(head!=null && head.val == duplicateVal)
                    head = head.next;
            }
        }
        return header.next;
    }
} 

Mistakes:

忘了update tail = tail.next