Remove Duplicates from Sorted List II

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：将前一个数与后一个数比较，如果不同且出现次数为1，则添加到结果里面。特别要注意最后一个结点：如果与前一个结点相同，则不必添加；如果与前一个结点不同，则必须要添加。运行时间14ms。思考：把line30~32为啥会出现wrong answer？

/**
 * 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 || !head->next) return head;
        
        ListNode *result = new ListNode(0);
        ListNode *preHead = result;
        int times = 1;
        while(head->next){
            if(head->val == head->next->val) times++;
            else{
                if(times == 1){
                    result->next = head;
                    result = result->next;
                }
                times = 1;
            }
            head = head->next;
        }
        if(times == 1){
            result->next = head;
            result = result->next;
        }
        result->next = NULL;
        
        return preHead->next;
    }
};