leetcode 82 82. 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.

Example 1:

Input: 1->2->3->3->4->4->5
Output: 1->2->5

Example 2:

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

思路就是把重复的结点的值设置为一个num  然后往后遍历 如果后面的值与这个相等 就删除这个结点；

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==nullptr)
            return nullptr;
        ListNode *dummy=new ListNode (0);
        ListNode *cur=dummy;
        dummy->next=head;
        while(cur->next&&cur->next->next)
        {
            if(cur->next->val==cur->next->next->val)
            {
                int num=cur->next->val;//重复的值进行 标记备份 如果下面的值有与它相等的就直接删除；
                while(cur->next&&cur->next->val==num)
                    cur->next=cur->next->next;
            }
            else 
                cur=cur->next;
        }
        return dummy->next;
        
    }
};