82. Remove Duplicates from Sorted List II

Medium

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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.

New Solution:

# 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:
            return head
        dummy = previous = ListNode()
        previous.next = head
        dummy.next = head
        while head and head.next:
            if head.val == head.next.val:
                while head and head.next and (head.val == head.next.val):
                    head = head.next
                previous.next = head.next
            else:
                previous = previous.next
            head = head.next
        return dummy.next

Old Solution

# 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 head is None:
            return
        new = head
        prev = None
        old = None
        if head.next is None:
            return head
        prev = new
        new = new.next
        repeat = False
        newHead = None
        while new:
            if new.val != prev.val:
                if not repeat:
                    if old is None:
                        old = prev
                        if newHead is None:
                            newHead = old
                    else:
                        old.next = prev
                        old = prev
                prev = new
                new = new.next
                repeat = False
            elif new.val == prev.val:
                repeat = True
                prev = new
                new = new.next
        if newHead is None and not repeat:
            return prev
        elif newHead is None:
            return None
        if not repeat:
            old.next = prev
        else:
            old.next = None
        return newHead