Finding the kth to last element of a singly linked list

I'm not sure if this code is industry-standard or not. Please review it and let me know if any improvement is required.

class Node(object):
   def __init__(self,data,next):
        self.data=data
        self.next=next

class getLinkRest():
       head = None
       tail= None

   def insert(self,data):  
      node=Node(data,None)
     if self.head is None:
       self.head=self.tail=node
     else:
       self.tail.next=node
     self.tail=node

 def showFromItem(self,data):
     curr_node=self.head
     flag=None
     while curr_node is not None:
          if curr_node.data == data or curr_node.data > data:
            print curr_node.data
            flag =1
            curr_node=curr_node.next
    if(flag is None):
       print "Not exist"

  s= getLinkRest()
  s.insert(1)
  s.insert(2)
  s.insert(3)
  s.insert(4)
  s.insert(5)
  s.showFromItem(3)
  s.showFromItem(7)

Fix the indentation

Before I say anything, let's reformat your code,
for everyone's sake,
because as you posted it, it just doesn't work,
because indentation is crucial in Python.
Please be more careful next time.

class Node(object):
    def __init__(self, data, next):
        self.data = data
        self.next = next

class getLinkRest():
    head = None
    tail = None

    def insert(self, data):
        node = Node(data, None)

        if self.head is None:
            self.head = self.tail = node
        else:
            self.tail.next = node
        self.tail = node

    def showFromItem(self, data):
        curr_node = self.head
        flag = None
        while curr_node is not None:
            if curr_node.data == data or curr_node.data > data:
                print curr_node.data
                flag = 1
            curr_node = curr_node.next

        if flag is None:
            print "Not exist"

s = getLinkRest()
s.insert(1)
s.insert(2)
s.insert(3)
s.insert(4)
s.insert(5)
s.showFromItem(3)
s.showFromItem(7)

Purpose of the code


Finding the kth to last element of a singly linked list

This is not what the posted code does.
The showFromItem function searches for an element with the specified value,
and prints that element and the rest of the elements until the end of the linked list that are bigger than the the specified value, or else it prints the text "Not exist".
For example:

s.insert(11)
s.insert(12)
s.insert(13)
s.insert(4)
s.insert(15)
s.showFromItem(13)

The above code will print 13 and 15.

That's a very unusual logic, and somehow I doubt that this was really your intention.

It's good to have a clear purpose, and to include a clear problem statement in a question.

Unused constructor argument

The next value of the constructor is never used.
So you could simplify it:

class Node(object):
    def __init__(self, data):
        self.data = data
        self.next = None

Instead of using flag with None and 1 as values,
it would be more natural to make it a boolean, like this:

found = False
    while curr_node is not None:
        if curr_node.data == data or curr_node.data > data:
            print(curr_node.data)
            found = True
        curr_node = curr_node.next

    if not found:
        print("Not exist")

You can write this condition shorter:

if curr_node.data == data or curr_node.data > data:

Like this:

if curr_node.data >= data:

StackExchange Code Review Q#98156, answer score: 7