82. 删除排序链表中的重复元素 II

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        dummy = ListNode(-1, head)
        cur = dummy
        while cur.next and cur.next.next:
            if cur.next.val == cur.next.next.val:
                val = cur.next.val
                while cur.next and cur.next.val == val:
                    cur.next = cur.next.next
            else:
                cur = cur.next
        return dummy.next

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode deleteDuplicates(ListNode head) {
        ListNode dummy = new ListNode(-1, head);
        ListNode cur = dummy;
        while (cur.next != null && cur.next.next != null) {
            if (cur.next.val == cur.next.next.val) {
                int val = cur.next.val;
                while (cur.next != null && cur.next.val == val) {
                    cur.next = cur.next.next;
                }
            } else {
                cur = cur.next;
            }
        }
        return dummy.next;
    }
}

/**
 * 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) {
        ListNode* dummy = new ListNode(-1, head);
        ListNode* cur = dummy;
        while (cur->next != nullptr && cur->next->next != nullptr) {
            if (cur->next->val == cur->next->next->val) {
                int val = cur->next->val;
                while (cur->next != nullptr && cur->next->val == val) {
                    cur->next = cur->next->next;
                }
            } else {
                cur = cur->next;
            }
        }
        return dummy->next;
    }
};