03 Jan 82. Remove Duplicates from Sorted List II
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.
Recursive
-
Idea
- Base case: list is empty or contains only one element –> No duplicates
- Recursive case: Assuming that the rest of the list is already done removing duplicates, we only need to consider the first part
- There’s no duplicate in the first part
- There’re duplicates in the first part
- There’s no duplicate in the first part
# 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]:
if not head or not head.next:
return head
if head.val != head.next.val:
head.next = self.deleteDuplicates(head.next)
return head
current = head
while current.next and current.next.val == current.val:
current = current.next
return self.deleteDuplicates(current.next)
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