## Description

https://leetcode.com/problems/remove-duplicates-from-an-unsorted-linked-list/

Given the `head`

of a linked list, find all the values that appear **more than once** in the list and delete the nodes that have any of those values.

Return *the linked list after the deletions.*

**Example 1:**

Input:head = [1,2,3,2]Output:[1,3]Explanation:2 appears twice in the linked list, so all 2's should be deleted. After deleting all 2's, we are left with [1,3].

**Example 2:**

Input:head = [2,1,1,2]Output:[]Explanation:2 and 1 both appear twice. All the elements should be deleted.

**Example 3:**

Input:head = [3,2,2,1,3,2,4]Output:[1,4]Explanation:3 appears twice and 2 appears three times. After deleting all 3's and 2's, we are left with [1,4].

**Constraints:**

- The number of nodes in the list is in the range
`[1, 10`

^{5}] `1 <= Node.val <= 10`

^{5}

## Explanation

We can first iterate the linked list to count the value occurrences. Then iterate again on the linked list, if the node value is not duplicated, append that value to a new linked list.

## 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 deleteDuplicatesUnsorted(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
new_dummy = ListNode(0)
new_head = new_dummy
while head != None:
if counter[head.val] > 1:
pass
else:
new_head.next = ListNode(head.val)
new_head = new_head.next
head = head.next
return new_dummy.next
```

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