Given a sorted linked list, delete all duplicates such that each element appear only once.

Example 1:

Input: 1->1->2
Output: 1->2

Example 2:

Input: 1->1->2->3->3
Output: 1->2->3

方法1：利用set，count()查找当前元素的个数，存在删除，不存在添加。使用两个指针记录前一个和当前节点。

/**
 * 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) {
        set<int> nodeValue;
        ListNode *pre=head, *cur=head;
        if(head==NULL) return head;
        cur=pre->next;
        nodeValue.insert(head->val);
        while(cur!=NULL){
            if(nodeValue.count(cur->val)==1){
                pre->next=cur->next;
            }else{
                nodeValue.insert(cur->val);
                pre=cur;
            }   
            cur=cur->next;
        }
        return head;
    }
};

tips:指针使用->访问元素，定义 nodeList *pre, *cur;