代码思路:首先判断链表是否不存在节点或只有一个节点,答案为真则返回原链表,若答案为假,则比较当前节点与下一节点的值,若两值相等则排除后者,不相等则节点指针后移,直至链表比较完成
class Solution:
def deleteDuplicates(self, head: ListNode) -> ListNode:
node=head
if node is None or node.next is None:
return head
while node.next:
if node.val==node.next.val:
node.next=node.next.next
else:
node=node.next
return head