## Description

https://leetcode.com/problems/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.

## Explanation

We can count the node value occurrences first and then remove those nodes with duplicated values.

## Python 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: ListNode) -> ListNode:
counter = {}
dummy = ListNode(0)
dummy.next = head
while head != None:
counter[head.val] = counter.get(head.val, 0) + 1
head = head.next
head = dummy.next
prev = dummy
while head != None:
if counter[head.val] > 1:
prev.next = head.next
else:
prev = head
head = head.next
return dummy.next
```

- Time Complexity: O(N).
- Space Complexity: O(N).

I found that solution very popular and helpful:

https://www.youtube.com/watch?v=oMK-uH-zAyQ&ab_channel=EricProgramming