力扣---删除排序链表中的重复元素 II

给定一个已排序的链表的头 head ， 删除原始链表中所有重复数字的节点，只留下不同的数字 。返回 已排序的链表 。

示例 1：

输入：head = [1,2,3,3,4,4,5]
输出：[1,2,5]

示例 2：

输入：head = [1,1,1,2,3]
输出：[2,3]

思路：三个指针，一次遍历，遇到重复节点删除即可

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode() : val(0), next(nullptr) {}
 *     ListNode(int x) : val(x), next(nullptr) {}
 *     ListNode(int x, ListNode *next) : val(x), next(next) {}
 * };
 */
 
//三个指针  一次遍历
 
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        if(head==nullptr){
            return head;
        }
 
        ListNode * pre = new ListNode(0,head);
        ListNode * start = pre;
        ListNode * end = pre;
        ListNode * end_pre = pre;
 
        start = start->next->next;
        end = end->next;
 
        int flag = 0;
        while(start!=nullptr){
            if(start->val == end->val){
                end->next = start->next;
                start = start->next;
                flag = 1;
            }
            else{
                if(flag==0){
                    end = end ->next;
                    start = start->next;   
                    end_pre = end_pre->next;                 
                }
                else{
                    end_pre->next = end_pre->next->next;
                    end = end->next;
                    start = start->next;
                    flag = 0;
                }
            }
        }
        if (flag==1){
            end_pre->next = end_pre->next->next;
            end = end->next;
        }
        return pre->next;
    }
};