[Force] 82. Delete button to sort the list of repeating elements II

First, the subject description:

Given a sorted list , delete all of the nodes contain repeated digits, leaving only the digital original list is not recurring.

Example 1:

输入: 1->2->3->3->4->4->5
输出: 1->2->5

Example 2:

输入: 1->1->1->2->3
输出: 2->3

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/remove-duplicates-from-sorted-list-ii
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

Second, the problem-solving ideas:

1, and traverse the list to find whether the element is repeated.
2, if not repeat, create a new list, add a new element to the current list.
3, if duplicate, delete duplicate nodes.
Given a sorted list: that the current list and orderly, repeating certain elements adjacent.

Third, Code Description:

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */

class Solution {
    public ListNode deleteDuplicates(ListNode head) {
    //创建一个新链表
     ListNode newHead=new ListNode(-1);
     ListNode newTail=newHead;
     ListNode cur=head;
     while(cur!=null){
     //针对某个引用进行 . 操作的时候, 必须保证该引用不能是 null
         if(cur.next!=null&&cur.val==cur.next.val){
         //当前元素是重复元素，查找重复元素区间。
             while(cur.next!=null&&cur.val==cur.next.val){
                 cur=cur.next;
             }
             //while循环结束后，cur指向重复元素中的最后一个位置。
             //cur向前再走一步，跳过了整个相同元素的区间。
             cur=cur.next;
         }
         else{
         //当前元素不是重复元素，插入到新链表中
             newTail.next=new ListNode(cur.val);
             newTail=newTail.next;
             cur=cur.next;
         }
     }
     return newHead.next;
    }
}