leetcode 82. Delete duplicate Element II in the sorting linked list

difficulty ： secondary
The frequency of ：77
subject ： Given the header of a sorted linked list head , Delete all nodes with duplicate numbers in the original linked list , Leave only different numbers . return Sorted linked list .

Their thinking ： One traverse
Be careful ：

• Traversal time , If the values of two nodes are equal , that cur The pointer moves
• most important of all , Not equal to the time of processing , But when they are not equal
  • When it's not equal , To judge ,pre and cur It's not a relationship
    - If it's still contextual , Then put the pre Move to cur
    - If it's not context , That explains. pre And then to cur（ Include cur） They are all equal , So directly put pre Connect to cur The next node of
• Finally, because it is traversal , So definitely cur To move
• To sum up ： Move when equal cur, When it's not equal pre and cur All move , At the end of a cycle, one after another
  Code ：

/** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode() {} * ListNode(int val) { this.val = val; } * ListNode(int val, ListNode next) { this.val = val; this.next = next; } * } */
class Solution {
    
    public ListNode deleteDuplicates(ListNode head) {
    
        ListNode dummyhead=new ListNode(-1);
        dummyhead.next=head;
        ListNode cur=head,pre=dummyhead;
        while(cur!=null){
    // Because it's traversal , So this condition is certain 
            while(cur.next!=null&&cur.val==cur.next.val){
       Because of the use of cur.next.val, therefore cur.next!=null
            // also cur.next!=null It has to be in the front , Otherwise, a null pointer exception will occur 
                cur=cur.next;
            }
            //pre and cur Connected to each other , You don't have to do much 
            if(pre.next==cur){
    
                pre=cur;
            }
            //pre and cur There are equal in the middle   also cur Equal to the middle 
            else{
    
                pre.next=cur.next;
            }
            cur=cur.next;
        }                
        return dummyhead.next;
    }
}