# 删除排序链表中的重复元素 II
存在一个按升序排列的链表，给你这个链表的头节点 head ，请你删除链表中所有存在数字重复情况的节点，只保留原始链表中 没有重复出现 的数字。
返回同样按升序排列的结果链表。
示例 1：
输入：head = [1,2,3,3,4,4,5]
输出：[1,2,5]
示例 2：
输入：head = [1,1,1,2,3]
输出：[2,3]
提示：
- 链表中节点数目在范围
[0, 300] 内 -100 <= Node.val <= 100- 题目数据保证链表已经按升序排列
以下错误的选项是？
## aop
### before
```cpp
#include
using namespace std;
struct ListNode
{
int val;
ListNode *next;
ListNode(int x) : val(x), next(NULL) {}
};
```
### after
```cpp
```
## 答案
```cpp
class Solution
{
public:
ListNode *deleteDuplicates(ListNode *head)
{
if (!head || !head->next)
return head;
auto dummy = new ListNode(-1);
dummy->next = head;
auto pre = dummy, cur = head;
while (cur && cur->next)
{
if (cur->val != cur->next->val)
{
pre = cur;
cur = cur->next;
}
else
{
while (cur->next && cur->val == cur->next->val)
{
cur = cur->next;
}
}
}
return dummy->next;
}
};
```
## 选项
### A
```cpp
class Solution
{
public:
ListNode *deleteDuplicates(ListNode *head)
{
if (!head || !head->next)
return head;
if (head->val == head->next->val)
{
while (head->next && head->val == head->next->val)
{
head = head->next;
}
return deleteDuplicates(head->next);
}
else
head->next = deleteDuplicates(head->next);
return head;
}
};
```
### B
```cpp
class Solution
{
public:
ListNode *deleteDuplicates(ListNode *head)
{
ListNode *new_head = new ListNode(0);
ListNode *cur = new_head;
ListNode *temp1 = head;
while (temp1)
{
ListNode *temp2 = temp1->next;
if (!temp2)
{
cur->next = temp1;
temp1 = temp1->next;
cur = cur->next;
}
else if (temp2 && temp1->val != temp2->val)
{
cur->next = temp1;
temp1 = temp1->next;
cur = cur->next;
}
else
{
while (temp2 && temp1->val == temp2->val)
{
temp2 = temp2->next;
}
temp1 = temp2;
}
}
cur->next = NULL;
return new_head->next;
}
};
```
### C
```cpp
class Solution
{
public:
ListNode *deleteDuplicates(ListNode *head)
{
if (head == NULL || head->next == NULL)
return head;
ListNode *L = new ListNode(0);
L->next = head;
ListNode *slow = L;
while (slow->next)
{
ListNode *fast = slow->next;
while (fast->next && fast->val == fast->next->val)
{
fast = fast->next;
}
if (slow->next == fast)
slow = slow->next;
else
{
slow->next = fast->next;
}
}
return L->next;
}
};
```