LC.82. Remove Duplicates from Sorted List II

https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/description/
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.

time: o(n) complexity: o(1)

 1 public ListNode deleteDuplicates(ListNode head) {
 2         if (head == null || head.next != null ) return head ;
 3         ListNode dummy = new ListNode(0) ;
 4         //注意,如果 CURR = HEAD 的话, 那第一个节点是没有办法删除掉的! 所以要灵活的运用CURR
 5         ListNode curr = dummy;
 6         dummy.next = head ;
 7         /*
 8         *   1->2->3->3->4->4->5, return 1->2->5.
 9         * c c  c-----c
10         * */
11         while (curr.next != null && curr.next.next != null){
12             if (curr.next.val == curr.next.next.val ){
13                 int val= curr.next.val;
14                 //凡事循环了,就要CHECK NPE
15                 while(curr.next != null && curr.next.val == val){
16                     curr.next = curr.next.next ;
17                 }
18             } else{
19                 curr = curr.next ;
20             }
21         }
22         return dummy.next ;
23     }
原文地址:https://www.cnblogs.com/davidnyc/p/8460959.html