用于从排序链表中删除所有重复项的Python程序

给定一个排序的链表，删除所有重复数字（所有出现）的节点，只留下在原始列表中出现一次的数字。
例子：

Input: 23->28->28->35->49->49->53->53
Output: 23->35

Input: 11->11->11->11->75->75
Output: empty List

请注意，这与从链接列表中删除重复项不同

这个想法是维护一个指向节点的指针（prev） ，该节点恰好位于我们正在检查重复的节点块之前。在第一个示例中，当我们检查节点 28 的重复项时，指针prev将指向 23。一旦我们到达值为 28 的最后一个重复节点（将其命名为当前指针），我们可以使 prev 节点的下一个字段成为当前的下一个并更新current=current.next 。这将删除具有重复项的值为 28 的节点块。

Python3

# Python3 implementation for the 
# above approach
  
# Creating node
class Node:
    def __init__(self, val):
        self.val = val
        self.next = None
class LinkedList:
    def __init__(self):
        self.head = None
          
    # Add node into beganing of linked list
    def push(self, new_data):
        new_node = Node(new_data)
        new_node.next = self.head
        self.head = new_node
        return new_node
          
    # Function to remove all occurrences
    # of duplicate elements
    def removeAllDuplicates(self, temp):
          
        # temp is head node of linkedlist
        curr = temp
  
        # print(' print something')
        head = prev = Node(None)
        head.next = curr
  
        # Here we use the same as we do in removing 
        # duplicates and the only extra thing is that
        # we need to remove all elements 
        # having duplicates that we did in 30-31
        while curr and curr.next:
              
            # until the current value and next value 
            # are same keep updating the current value
            if curr.val == curr.next.val:
                while(curr and curr.next and 
                      curr.val == curr.next.val):
                    curr = curr.next
                      
                    # still one of duplicate values left
                    # so point prec to curr
                curr = curr.next
                prev.next = curr
            else:
                prev = prev.next
                curr = curr.next
        return head.next
          
    # for print the linkedlist
    def printList(self):
        temp1 = self.head
        while temp1 is not None:
            print(temp1.val, end = " ")
            temp1 = temp1.next
              
    # For getting head of linkedlist
    def get_head(self):
        return self.head
  
# Driver Code
if __name__=='__main__':
    llist = LinkedList()
    llist.push(53)
    llist.push(53)
    llist.push(49)
    llist.push(49)
    llist.push(35)
    llist.push(28)
    llist.push(28)
    llist.push(23)
      
    print('Created linked list is:')
    llist.printList()
    print(
    'Linked list after deletion of duplicates:')
    head1 = llist.get_head()
    #print(head1)
    llist.removeAllDuplicates(head1)
    llist.printList()
          
# This code is contributed by PRAVEEN KUMAR(IIIT KALYANI)

输出：

List before removal of duplicates
23 28 28 35 49 49 53 53 
List after removal of duplicates
23 35 

时间复杂度： O(n)
有关详细信息，请参阅有关从排序的链接列表中删除所有重复项的完整文章！