# 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
以下程序实现了这一功能，请你填补空白处内容：
```cpp
#include
#include
struct ListNode
{
int val;
struct ListNode *next;
};
struct ListNode *deleteDuplicates(struct ListNode *head)
{
struct ListNode dummy;
struct ListNode *p, *q, *prev;
prev = &dummy;
dummy.next = head;
p = q = head;
while (p != NULL)
{
_____________________
}
return dummy.next;
}
int main(int argc, char **argv)
{
int i;
struct ListNode *head = NULL;
struct ListNode *prev = NULL;
struct ListNode *p;
for (i = 0; i < argc - 1; i++)
{
p = malloc(sizeof(*p));
p->val = atoi(argv[i + 1]);
p->next = NULL;
if (head == NULL)
{
head = p;
prev = head;
}
else
{
prev->next = p;
prev = p;
}
}
p = deleteDuplicates(head);
while (p != NULL)
{
printf("%d ", p->val);
p = p->next;
}
printf("\n");
return 0;
}
```
## template
```cpp
#include
#include
struct ListNode
{
int val;
struct ListNode *next;
};
struct ListNode *deleteDuplicates(struct ListNode *head)
{
struct ListNode dummy;
struct ListNode *p, *q, *prev;
prev = &dummy;
dummy.next = head;
p = q = head;
while (p != NULL)
{
while (q != NULL && q->val == p->val)
{
q = q->next;
}
if (p->next == q)
{
prev = p;
}
else
{
prev->next = q;
}
p = q;
}
return dummy.next;
}
int main(int argc, char **argv)
{
int i;
struct ListNode *head = NULL;
struct ListNode *prev = NULL;
struct ListNode *p;
for (i = 0; i < argc - 1; i++)
{
p = malloc(sizeof(*p));
p->val = atoi(argv[i + 1]);
p->next = NULL;
if (head == NULL)
{
head = p;
prev = head;
}
else
{
prev->next = p;
prev = p;
}
}
p = deleteDuplicates(head);
while (p != NULL)
{
printf("%d ", p->val);
p = p->next;
}
printf("\n");
return 0;
}
```
## 答案
```cpp
while (q != NULL && q->val == p->val)
{
q = q->next;
}
if (p->next == q)
{
prev = p;
}
else
{
prev->next = q;
}
p = q;
```
## 选项
### A
```cpp
while (q != NULL && q->val == p->val)
{
q = q->next;
}
if (p->next == q)
{
prev = p;
}
else
{
prev->next = q;
}
```
### B
```cpp
while (q != NULL && q->val == p->val)
{
q = q->next;
}
if (p->next == q)
{
prev = p;
}
p = q;
```
### C
```cpp
while (q != NULL && q->val == p->val)
{
q = q->next;
}
if (p->next == q)
{
prev->next = q;
}
p = q;
```