## The Problem

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 besortedin ascending order.

## My Tests

```
import pytest
from typing import List
from .util import ListNode, toListNode, toList
from .Day5 import Solution
s = Solution()
@pytest.mark.parametrize(
"input,expected",
[
([1, 2, 3, 3, 4, 4, 5], [1, 2, 5]),
([1, 2, 3, 3, 3, 4, 4, 5], [1, 2, 5]),
([1, 1, 1, 2, 3], [2, 3]),
([2, 3, 4], [2, 3, 4]),
([1, 1], []),
([0], [0]),
([], []),
([1, 1, 2, 2], []),
],
)
def test_remove_duplicates(input, expected):
assert toList(s.deleteDuplicates(toListNode(input))) == expected
```

## My Solution

```
class Solution:
def deleteDuplicates(self, head: ListNode) -> ListNode:
holder = ListNode(0, head)
dummy_head = holder
next_node = head
while next_node:
if next_node.next is not None and next_node.next.val == next_node.val:
while (
next_node.next is not None and next_node.val == next_node.next.val
):
next_node = next_node.next
dummy_head.next = next_node.next
else:
dummy_head = dummy_head.next
next_node = next_node.next
return holder.next
```

## Analysis

## My Commentary

This was another one I thought would be super easy but I got confused a bunch of times trying to figure out the pointers. I even messed up my first attempt because I didn't read the question correctly and thought I just had to remove duplicates without removing all instances of a number that had a match.

I did think initially I would need to keep pointers for the current head and the next list item. I assumed all I would have to do is increment the `next`

pointer while there was a match. As soon as we didn't have a match, bump the `head.next`

to `next`

. Of course, we would then need another pointer for `head.next`

to carry on. I ended up with a bunch of if statements and a mess of code.

It took me a while to realise the easiest thing to do is create a `holder`

pointer that would easily allow me to increment `head`

(when there are duplicates at the beginning) and then we can set up a dummy head pointer as well as a `next_node`

pointer.

So then we check for duplicates, increment the `next_node`

pointer for all those duplicates. Then we push the `dummy_head`

pointer past the duplicates which we can safely do because we kept the original head in the `holder`

.

If we don't see a duplicate we just increment the `dummy_head`

pointer to the next node in

the list.

I feel like if I got this in an interview yesterday I would have been in trouble even though the solution seems so obvious now.

## Discussion (0)