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:
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def deldup(self, head: Optional[ListNode]) -> Optional[ListNode]:
if not head:
return head
if head.val in self.dups or (head.next and head.val == head.next.val):
self.dups.add(head.val)
head = self.deldup(head.next)
else:
head.next = self.deldup(head.next)
return head
def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
self.dups = set()
head = self.deldup(head)
return head
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