<p>Given the <code>head</code> of a sorted linked list, <em>delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list</em>. Return <em>the linked list <strong>sorted</strong> as well</em>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/linkedlist1.jpg" style="width: 500px; height: 142px;" />
<pre>
<strong>Input:</strong> head = [1,2,3,3,4,4,5]
<strong>Output:</strong> [1,2,5]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2021/01/04/linkedlist2.jpg" style="width: 500px; height: 205px;" />
<pre>
<strong>Input:</strong> head = [1,1,1,2,3]
<strong>Output:</strong> [2,3]
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li>The number of nodes in the list is in the range <code>[0, 300]</code>.</li>
	<li><code>-100 &lt;= Node.val &lt;= 100</code></li>
	<li>The list is guaranteed to be <strong>sorted</strong> in ascending order.</li>
</ul>

# 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 or not head.next:
            return head

        output=ListNode(0,head)
        pre=output
        while head:
            if head.next and head.val==head.next.val:
                while head.next and head.val==head.next.val:
                    head=head.next
                pre.next=head.next
            else:
                pre=pre.next
            head=head.next
        # return output.next


        return output.next
16500