力扣82. 删除排序链表中的重复元素 II

 暴力解：

/**
 * 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) {}
 * };
 */
class Solution {
public:
    ListNode* deleteDuplicates(ListNode* head) {
        unordered_map<int, bool> record;
        ListNode *vhead = new ListNode(0, head);
        if (head && head -> next) {
            ListNode *pre = vhead, *cur = head, *nxt = head -> next;
            while(nxt){ //遍历以尾指针有效为边界
                while (cur -> val == nxt -> val) { //删除后续重复元素
                    record[cur -> val] = true;
                    cur -> next = nxt -> next;
                    nxt = cur -> next;
                    if (nxt == nullptr){
                        break;
                    }
                }
                if (record[cur -> val]) { //删除最早的重复元素
                    pre -> next = cur -> next;
                    cur = pre -> next;
                    if (cur)
                        nxt = cur -> next;
                } else { //正常更新指针
                    pre = cur;
                    cur = nxt;
                    nxt = nxt -> next;
                }
            }

        }
        return vhead -> next;
    }
};