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
Python3
# 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]:
curr = res = ListNode(-1, head)
while curr:
if curr.next and curr.next.next and curr.next.val == curr.next.next.val:
v = curr.next.val
p = curr.next
while p and p.val == v:
p = p.next
curr.next = p
else:
curr = curr.next
return res.next