Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well. Example 1:Input: head = [1,1,2]Output: [1,2]Example 2:Input: head = [1,1,2,3,3]Output: [1,2,3] Constraints:The number of nodes in the list is in the range [0, 300].-100 <= Node.val <= 100The list is guaranteed to be sorted in ascending order.

Sure, here is a Python solution for the problem:

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

def deleteDuplicates(head):
    current = head
    while current is not None and current.next is not None:
        if current.next.val == current.val:
            current.next = current.next.next
        else:
            current = current.next
    return head

Here is a step by step explanation of the solution:

1. Define a class ListNode to represent each node in the linked list. Each node has a value val and a pointer next to the next node.

2. Define a function deleteDuplicates to remove duplicates from the linked list. The function takes the head of the linked list as input.

3. Inside the function, initialize a variable current to the head of the linked list.

4. Start a while loop that continues as long as current is not None and current.next is not None. This is to ensure that we don't access None values.

5. Inside the loop, check if the value of the next node is the same as the value of the current node. If it is, set the next node of the current node to the node after the next node, effectively skipping over the duplicate node.

6. If the value of the next node is not the same as the value of the current node, move current to the next node.

7. After the