题目
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.
分析
已排序的链表，删除含有重复数字的节点，只保留独特的值。因此可以依次遍历链表，找到重复的，就直接略去。不重复的就需要保留。特别注意的是链表首尾节点的处理，容易出错。
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* struct ListNode *next;
* };
*/
struct ListNode* deleteDuplicates(struct ListNode* head) {
struct ListNode *p=head,*q=head;
bool firstnode=true;
if(p==NULL)return head;
while(p!=NULL)
{
int num=0;
while(p->next!=NULL && p->val== p->next->val)
{
num++;
p=p->next;
}
if(p->next==NULL)
{
if(firstnode==true)
{
if(num==0)
head=p;
else
head=NULL;
}
else
{
if(num==0)
q->next=p;
else
q->next=NULL;
}
break;
}
if(num>0)
{
p=p->next;
}
else
{
if(firstnode==true)
{
head=p;
q=p;
p=p->next;
firstnode=false;
}
else
{
q->next=p;
q=p;
p=p->next;
}
}
}
return head;
}
