Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,
Given 1->1->2, return 1->2.
Given 1->1->2->3->3, return 1->2->3.

<span style="font-weight: normal;">/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head==null || head.next==null) return head;
        ListNode prev = head, cur = head.next;
        while(cur!=null){
            if(prev.val==cur.val){//remove cur from the list
                prev.next=cur.next; 
                cur=cur.next;
            }
            else{
                prev=prev.next;
                cur=cur.next;
            }
        }
        return head;
        
    }
}</span>

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.

<span style="font-size:18px;">/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        if(head==null || head.next==null) return head;
        ListNode newHead = new ListNode(0);//create a new empty head, the value of the new head will not be counted in
        newHead.next=head;
        ListNode prev = newHead, cur=head;
        while(cur!=null&&cur.next!=null){
            if(cur.val!=cur.next.val){
                prev=prev.next;
                cur=cur.next;
            }else{
                while(cur.next!=null&&cur.next.val==cur.val)    cur=cur.next;
                //remove nodes from prev.next to cur
                prev.next=cur.next;
                cur=cur.next;
            }
        }
        return newHead.next;
    }
}</span>