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题目链接：力扣（LeetCode）官网 - 全球极客挚爱的技术成长平台

 解题思路：设置一个新的哑元节点result，作为头节点，将head中不重复地节点依次链接到哑元节点后面，最后返回result.next

1. 初始值：
   1. result = new ListNode();
   2. prev = result
   3. current = head
   4. cnt = 0
2. 如果current != null，则循环执行：
   1. 如果 current.next!=null && current.next.val == current.val：说明节点重复
      1. 令current = current.next
      2. cnt++：重复节点的数量加1
      3. 如果 cnt>1 &&(current.next==null || current.next.val != current.val)：
         1. 此时说明有重复的节点，并且current已经到达最后一个重复的节点，但是后面的节点还有可能会出现重复，继续遍历后面的节点，
         2. current = current.next。
         3. cnt=0，重新计数
      4. continue，遍历下一个节点
   2. prev.next = current：将当前不重复的节点链接到新链表中。
   3. prev = current：更新前驱
   4. curent = current.next
   5. prev.next = null：因为current后面可能还会有重复的节点，所以prev的后继指向null，断开与current后面节点的链接

AC代码：

/*** 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) {if (head == null || head.next == null) {return head;}ListNode result = new ListNode();ListNode prev = result;ListNode current = head;int cnt = 0;while (current != null) {if (current.next != null && current.next.val == current.val) {current = current.next;cnt++;if (cnt > 0 && (current.next == null || current.next.val != current.val)) {current = current.next;cnt = 0;}continue;}prev.next = current;prev = current;current = current.next;prev.next=null;}return result.next;}
}

解法二：在头节点前添加一个哑元节点，初始时将current指向哑元节点，如果后面节点有重复的，就一直令current.next = current.next.next，丢弃中间重复的节点current.next，否则令current = current.next，指向下一个不重复的节点

AC代码：

public ListNode deleteDuplicates(ListNode head) {if (head == null || head.next == null) {return head;}ListNode dummy = new ListNode(0, head);ListNode current = dummy;while (current.next != null && current.next.next != null) {if (current.next.val == current.next.next.val) {int value = current.next.val;while (current.next != null && current.next.val == value) {current.next = current.next.next;}} else {current = current.next;}}return dummy.next;}