Question：

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.

Algorithm：

在头结点前申请一个内存，用两个节点指向重复的头和尾

Accepted Code：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(head==NULL||head->next==NULL)
            return head;
        ListNode* pre=new ListNode(0);         //在头节点前插一个节点,可以不用判断头节点和后面的值相等
        pre->next=head;
        ListNode* p=pre;

        while(p->next!=NULL)
        {
            ListNode* first=p->next;         //重复元素首节点
            ListNode* last=first->next;      //重复元素尾节点
            int count=0;
            while(last!=NULL&&first->val==last->val)
            {
                last=last->next;
                count++;
            }
            if(count>0)
                p->next=last;
            else
                p=p->next;
        }
        return pre->next;
    }
};