Description

Given the head of a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list. Return the linked list sorted as well.

Example 1:

Input: head = [1,2,3,3,4,4,5]
Output: [1,2,5]

Example 2:

Input: head = [1,1,1,2,3]
Output: [2,3]

Constraints:

The number of nodes in the list is in the range [0, 300].
-100 <= Node.val <= 100
The list is guaranteed to be sorted in ascending order.

Solution

Use a prev to record the previous node, and if the current node is duplicated by the next node, delete them. Otherwise move prev forward.

Time complexity: $o(n)$
Space complexity: $o(1)$

Code

# 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: Optional[ListNode]) -> Optional[ListNode]:
        ret_head = ListNode(-1)
        ret_head.next = head
        prev, p = ret_head, ret_head.next
        while p and p.next:
            pn = p.next
            need_delete = False
            while pn and pn.val == p.val:
                need_delete = True
                pn = pn.next
            if need_delete:
                prev.next = pn
                p = prev.next
            else:
                prev, p = prev.next, p.next
        return ret_head.next