题目：
给定一个排序链表，删除所有含有重复数字的节点，只保留原始链表中 没有重复出现 的数字。

示例 1:

输入: 1->2->3->3->4->4->5
输出: 1->2->5
示例 2:

输入: 1->1->1->2->3
输出: 2->3

链接：https://leetcode-cn.com/problems/remove-duplicates-from-sorted-list-ii

思路：
1、定义一个头结点。遍历链表，如果当前节点存在，并且当前节点的下一节点存在并且相等，则将其继续往后循环，直至找到节点值不相等的下一节点，否则就保留当前节点值的结果

Python代码：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        if not head or not head.next:
            return head
        ret = ListNode(0)
        guard = ret

        while head and head.next:
            if head.val == head.next.val:
                while head.next and head.val == head.next.val:
                    head = head.next
                ret.next = head.next
            else:
                ret.next = head
                ret = ret.next
            head = head.next
        return guard.next

C++代码：

/**
 * 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 || head->next==nullptr){
            return head;
        }

        ListNode* ret = new ListNode(0);
        ListNode* guard = ret;

        while (head!=nullptr && head->next!=nullptr){
            if (head->val == head->next->val){
                while (head->next!=nullptr && head->val == head->next->val){
                    head = head->next;
                }
                ret->next = head->next;
            }else{
                ret->next = head;
                ret = ret->next;
            }
            head = head->next;
        }
        return guard->next;
    }
};