题目链接：Remove Duplicates from Sorted List II

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.

这道题的要求是在有序链表中删除有重复数字的节点，留下没有重复的节点。

又是先出现的II题目，是Remove Duplicates from Sorted List的扩展，其实这俩题也都差不多，基础的链表操作。

思路就是找到重复元素，删除即可。如果两个连续节点的数字一样，则循环检测，删除重复的元素，不过最后还需要删除该重复节点一下，因为要求不保留重复节点。

时间复杂度：O(n)

空间复杂度：O(1)

 1 class Solution
 2 {
 3 public:
 4     ListNode *deleteDuplicates(ListNode *head)
 5     {
 6         ListNode *h = new ListNode(0), *p = h;
 7         h -> next = head;
 8         
 9         while(p -> next != NULL)
10         {
11             if(p -> next -> next != NULL && 
12                p -> next -> val == p -> next -> next -> val)
13             {
14                 while(p -> next -> next != NULL && 
15                       p -> next -> val == p -> next -> next -> val)
16                     p -> next = p -> next -> next;
17                 p -> next = p -> next -> next;
18             }
19             else
20                 p = p -> next;
21         }
22         
23         return h -> next;
24     }
25 };