考虑到头两个可能就是重复，申请使用一个dummy节点，
1 ，当遇见前后重复的节点就删除当前节点，并记录下重复节点的val.
2, 下次遍历如果遇见当前节点的值和dup val一样，就删除当前节点
*没有释放被删除节点，下次改进
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* deleteDuplicates(struct ListNode* head) {
if(head == NULL)
return NULL;
struct ListNode *dummy = calloc(1, sizeof(struct ListNode));
dummy->next = head;
struct ListNode * prev, * curr, *next, *dup;
struct ListNode *tmp;
prev = dummy;
curr = dummy->next;
next = curr->next;
dup = NULL;
while(curr){
if(dup && curr->val == dup->val){
tmp = curr;
curr = curr->next;
free(tmp);
if(curr)
next = curr->next;
else
next = NULL;
prev->next = curr;
}else if((next && next->val == curr->val)){
//delete next
prev->next = next->next;
//mark duplicate value
if(dup == NULL)
dup = calloc(1,sizeof(ListNode));
dup->val = next->val;
tmp = next;
free(tmp);
curr = prev->next;
if(curr)
next = curr->next;
else
next = NULL;
}else{
curr = curr->next;
if(curr)
next = curr->next;
else
next = NULL;
prev = prev->next;
}
}
struct ListNode * tmp2 = dummy->next;
free(dummy);
return tmp2;
}
