Suppose the code puzzle.extensions(self) has already been defined, and it will return a list that puzzle's available solutions but without the determination if it is solved. Also puzzle.is_solved(self) has been defined and it will determine whether this solution is solved. Here is the code that I need to write, I also do some incorrect works.
def depth_first_solve(puzzle):
"""
Return a path from PuzzleNode(puzzle) to a PuzzleNode containing
a solution, with each child containing an extension of the puzzle
in its parent. Return None if this is not possible.
#type puzzle: Puzzle
#rtype: PuzzleNode
"""
stack = [puzzle]
while stack:
k = stack.pop()
for puzzle1 in puzzle.extensions():
if not puzzle1.is_solved():
stack+=[k,puzzle1]
if puzzle1.is_solved():
p = stack.pop()
end_node = PuzzleNode(k,None, p)
k = stack.pop()
last_node = PuzzleNode(p,end_node,k)
while stack:
k = p
p = stack.pop()
cur_node = PuzzleNode(k, last_node, p)
last_node = cur_node
return cur_node
def __init__(self, puzzle=None, children=None, parent=None):
"""
Create a new puzzle node self with configuration puzzle.
#type self: PuzzleNode
#type puzzle: Puzzle | None
#type children: list[PuzzleNode]
#type parent: PuzzleNode | None
#rtype: None
"""
self.puzzle, self.parent = puzzle, parent
if children is None:
self.children = []
else:
self.children = children[:]
Well, I run these module in puzzle, and it always waiting for results and nothing happens, so could anyone tell me that where I got it wrong?
I think there are a very large number of issues with this code. To start with, you're always iterating on puzzle.extensions(), rather than on the extensions of the k node you've just popped off the stack. I suspect this is why you're getting an infinite loop, since the same nodes keep getting pushed onto the stack over and over (and ignored by the rest of the code).
I'm not really sure why you're adding k back to the stack though with stack+=[k,puzzle1]. I'm pretty sure you just want stack.append(puzzle1) there, unless you're trying something really subtle that I don't understand.
Related
I am doing/learning data structures and algorithms and I came across the following code:
class BinaryTree:
def __init__(self, root_data):
self.data = root_data
self.left_child = None
self.right_child = None
def inorder_iterative(self):
inorder_list = []
return inorder_list
def get_right_child(self):
return self.right_child
def get_left_child(self):
return self.left_child
def set_root_val(self, obj):
self.data = obj
def get_root_val(self):
return self.data
def preorder_iterative(self):
pre_ordered_list = [] #use this as result
stack = []
stack.append(self)
while len(stack)>0:
node = stack.pop() #should return a value
pre_ordered_list.append(node.get_root_val())
if node.right_child:
stack.append(node.get_right_child())
if node.left_child:
stack.append(node.get_left_child())
return pre_ordered_list
bn = BinaryTree(1)
bn.left_child = BinaryTree(2)
bn.right_child = BinaryTree(3)
print (bn.preorder_iterative())
I am very lost about stack.append(self) part. I am not sure what is the point of having this line and I don't fully understand the concept of .append(self). I have learnt that self represents the instance of the class.
The purpose of the stack is to simulate recursion.
The initial value placed on the stack is the tree itself (in the form of its root node). Its value is retrieved, and then each subtree is placed on the stack. On the next iteration of the loop, the left child is removed, processed, and replaced with its children, if any. The loop continues as long as there is anything on the stack to process. Once everything on the left side of the tree as been processed, you'll finally start on the right child (placed the stack way at the beginning of the loop).
Compare to a recursive version:
def preorder_recursive(self):
result = [self.get_root_val()]
if node.left_child:
result.extend(node.left_child.preorder_recursive())
if node.right_child:
result.extend(node.right_child.preorder_recursive())
return result
Each recursive call essentially puts self on a stack, allowing the left child (and its descendants) to be processed before eventually returning to the root and moving to its right child.
I'm trying to implement an iterator class for not-necessarily-binary trees in Python. After the iterator is constructed with a tree's root node, its next() function can be called repeatedly to traverse the tree in depth-first order (e.g., this order), finally returning None when there are no nodes left.
Here is the basic Node class for a tree:
class Node(object):
def __init__(self, title, children=None):
self.title = title
self.children = children or []
self.visited = False
def __str__(self):
return self.title
As you can see above, I introduced a visited property to the nodes for my first approach, since I didn't see a way around it. With that extra measure of state, the Iterator class looks like this:
class Iterator(object):
def __init__(self, root):
self.stack = []
self.current = root
def next(self):
if self.current is None:
return None
self.stack.append(self.current)
self.current.visited = True
# Root case
if len(self.stack) == 1:
return self.current
while self.stack:
self.current = self.stack[-1]
for child in self.current.children:
if not child.visited:
self.current = child
return child
self.stack.pop()
This is all well and good, but I want to get rid of the need for the visited property, without resorting to recursion or any other alterations to the Node class.
All the state I need should be taken care of in the iterator, but I'm at a loss about how that can be done. Keeping a visited list for the whole tree is non-scalable and out of the question, so there must be a clever way to use the stack.
What especially confuses me is this--since the next() function, of course, returns, how can I remember where I've been without marking anything or using excess storage? Intuitively, I think of looping over children, but that logic is broken/forgotten when the next() function returns!
UPDATE - Here is a small test:
tree = Node(
'A', [
Node('B', [
Node('C', [
Node('D')
]),
Node('E'),
]),
Node('F'),
Node('G'),
])
iter = Iterator(tree)
out = object()
while out:
out = iter.next()
print out
If you really must avoid recursion, this iterator works:
from collections import deque
def node_depth_first_iter(node):
stack = deque([node])
while stack:
# Pop out the first element in the stack
node = stack.popleft()
yield node
# push children onto the front of the stack.
# Note that with a deque.extendleft, the first on in is the last
# one out, so we need to push them in reverse order.
stack.extendleft(reversed(node.children))
With that said, I think that you're thinking about this too hard. A good-ole' (recursive) generator also does the trick:
class Node(object):
def __init__(self, title, children=None):
self.title = title
self.children = children or []
def __str__(self):
return self.title
def __iter__(self):
yield self
for child in self.children:
for node in child:
yield node
both of these pass your tests:
expected = ['A', 'B', 'C', 'D', 'E', 'F', 'G']
# Test recursive generator using Node.__iter__
assert [str(n) for n in tree] == expected
# test non-recursive Iterator
assert [str(n) for n in node_depth_first_iter(tree)] == expected
and you can easily make Node.__iter__ use the non-recursive form if you prefer:
def __iter__(self):
return node_depth_first_iter(self)
That could still potentially hold every label, though. I want the
iterator to keep only a subset of the tree at a time.
But you already are holding everything. Remember that an object is essentially a dictionary with an entry for each attribute. Having self.visited = False in the __init__ of Node means you are storing a redundant "visited" key and False value for every single Node object no matter what. A set, at least, also has the potential of not holding every single node ID. Try this:
class Iterator(object):
def __init__(self, root):
self.visited_ids = set()
...
def next(self):
...
#self.current.visited = True
self.visited_ids.add(id(self.current))
...
#if not child.visited:
if id(child) not in self.visited_ids:
Looking up the ID in the set should be just as fast as accessing a node's attribute. The only way this can be more wasteful than your solution is the overhead of the set object itself (not its elements), which is only a concern if you have multiple concurrent iterators (which you obviously don't, otherwise the node visited attribute couldn't be useful to you).
I've been trying to create a simple linked list implementation in Python as a code exercise and, although I have most of the stuff working (inserting, removing, pretty print, swapping the content of two nodes), I've been stuck on swapping two nodes for a few days.
I've looked around on the internet and most people seem to recommend deleting/inserting the nodes or swapping the data. Both are very fine and functional options but I wanted to challenge myself and see if I could swap the nodes the "correct" way.
Ideally I would like to have a generic function that can handle all edge cases (moving to the begin, end and swapping random nodes). This has proven to be way more challenging than I expected.
I've experimented a bit with pen and paper and search around and I found the following discussion and example implementation:
Discussion about swapping in C
Example implementation in C
The issue I run into is that my node1.next and my node2.prev are swapped and that my node2.next refers to itself and not to the next node in the list.
The comment on the page of the example implementation specifically addresses this problem and mentions that it should not happen with his implementation.
I can't seem to figure out what I've done wrong. I guess I could "cheat" and force them to take the correct values at the end but that gives a lot of problems when the node is the first/last.
__author__ = 'laurens'
from django.core.exceptions import ObjectDoesNotExist
import logging
import copy
logging.basicConfig(level=logging.DEBUG)
logger = logging.getLogger(__name__)
class DoublyLinkedList(object):
"""This class implements a basic doubly linked list in Django, it depends
on a Django model with the following field:
id : PK
prev: integer previous node
data_field: Foreign key
next: integer next node
the prev and next fields don't need to be self-referencing
When instantiating the class you have to link this class to a Django model
and specify a data field. The data field can link to a foreign key
or contain data
"""
def __init__(self, doubly_linked_list_model, data_field):
self.doubly_linked_list_model = doubly_linked_list_model
self.data_field = data_field
def get_node_from_node_id(self, node_id=None):
"""This function returns the node associated with a certain node_id"""
if node_id is None:
node = None
else:
try:
node = self.doubly_linked_list_model.get(id=node_id)
except ObjectDoesNotExist:
node = None
return node
#staticmethod
def _update_node(node, prev=None, next=None):
node.prev = prev
node.next = next
logger.debug('updating node: %s', node.id)
logger.debug('node.prev = %s', node.prev)
logger.debug('node.next = %s', node.next)
try:
node.save()
except Exception as e: #Todo: specify this
logger.debug('Error saving node: %s', node.id)
def move_node(self, node1=None, node2=None):
"""
This function swaps the position of node1 with the position of node2
"""
#swapping two nodes!
logger.debug('Swapping two random nodes!: %s, %s', node1.id, node2.id)
# Swapping next nodes
logger.debug('Swapping next node')
tmp = copy.deepcopy(node1.next)
self._update_node(node=node1,
prev=node1.prev,
next=node2.next)
#Todo: Check if tmp changes or is stored as a copy
self._update_node(node=node2,
prev=node2.prev,
next=tmp)
if node1.next is not None:
logger.debug('Connect the next node to node 1')
node_next = self.get_node_from_node_id(node1.next)
self._update_node(node=node_next,
prev=node1.id,
next=node_next.next)
if node2.next is not None:
logger.debug('Connect the next node to node 2')
node_next = self.get_node_from_node_id(node2.next)
self._update_node(node=node_next,
prev=node2.id,
next=node_next.next)
logger.debug('Swap prev nodes')
tmp = copy.deepcopy(node1.prev)
self._update_node(node=node1,
prev=node2.prev,
next=node1.next)
self._update_node(node=node2,
prev=tmp,
next=node2.next)
# Connect the node before node1 to node 1
if node1.prev is not None:
logger.debug('Connect the prev to node 1')
node_prev = self.get_node_from_node_id(node1.prev)
self._update_node(node=node_prev,
prev=node_prev.prev,
next=node1.id)
# Connect the node before node2 to node 2
if node2.prev is not None:
logger.debug('Connect the prev to node 2')
node_prev = self.get_node_from_node_id(node2.prev)
self._update_node(node=node_prev,
prev=node_prev.prev,
next=node2.id)
The _update_node function does nothing more than taking my input and committing it to the database; it can handle None values.
get_node_from_node_id takes an integer as input and returns the node object associated with it. I use it so that I don't have to work with self-referencing foreign keys (is that the correct term?) in the database, for now I would like to continue working this way. Once I have this working I'll move on to fixing it in the database in the correct way.
tip: I get answers much more quickly when I provide a minimal, complete, verifiable example (MCVE), also known as a short, self-contained, compilable example (SSCCE).
Your example code fails to demonstrate the problem, making it impossible for us to help you.
Please make it easy for us to help you.
I ran your example code, but I didn't see any output.
The issue I run into is that ... that my node2.next refers to itself and
not to the next node in the list.
Why is node2.next referring to node2 a problem?
As far as I can tell, the part of the code you gave us so far works fine.
Some of the most difficult debugging sessions I've ever had ended only when I realized that everything was actually working correctly, that the "bug" I thought I was hunting didn't actually exist.
Yes, there is an intermediate step halfway through the algorithm where a node refers to itself, which may seem obviously wrong.
But then the second half of the algorithm makes further changes.
By the time the algorithm finishes,
the "next" chain correctly runs all the way from the beginning to the end, and
the "forward" chain correctly runs in the reverse order as the "next" chain, from the end to the beginning, and
the order of those chains is similar to the original order, except that node1 and node2 have swapped logical position (but are still in the same physical position).
Every node points to 2 other nodes (or to the NULL node), never to itself.
Isn't that what you wanted?
When I ran the following test script, I get lots of output --
but I don't see any problems in the output.
(I used a simple array with integer indexes rather than true pointers or references to make the code shorter and easier to debug compared to a SQL database).
Would you mind pointing out which particular line of output is not what you expected, and spelling out what you expected that line of output to actually say?
#!/usr/bin/env python
# https://stackoverflow.com/questions/24610889/trying-to-write-a-swap-position-function-for-a-doubly-linked-list
# Is this the shortest possible program that exhibits the bug?
# Before running this probram, you may need to install
# sudo apt-get install python-django
# 2015-03-12: David added some test scaffolding
# 2014-07-07: Laurens posted to StackOverflow
__author__ = 'laurens'
from django.core.exceptions import ObjectDoesNotExist
import logging
import copy
logging.basicConfig(level=logging.DEBUG)
logger = logging.getLogger(__name__)
class DoublyLinkedList(object):
"""This class implements a basic doubly linked list in Django, it depends
on a Django model with the following field:
id : PK
prev: integer previous node
data_field: Foreign key
next: integer next node
the prev and next fields don't need to be self-referencing
When instantiating the class you have to link this class to a Django model
and specify a data field. The data field can link to a foreign key
or contain data
"""
def __init__(self, doubly_linked_list_model, data_field):
self.doubly_linked_list_model = doubly_linked_list_model
self.data_field = data_field
def get_node_from_node_id(self, node_id=None):
"""This function returns the node associated with a certain node_id"""
if node_id is None:
node = None
else:
try:
node = self.doubly_linked_list_model.get(id=node_id)
except ObjectDoesNotExist:
node = None
return node
#staticmethod
def _update_node(node, prev=None, next=None):
node.prev = prev
node.next = next
logger.debug('updating node: %s', node.id)
logger.debug('node.prev = %s', node.prev)
logger.debug('node.next = %s', node.next)
try:
node.save()
except Exception as e: #Todo: specify this
logger.debug('Error saving node: %s', node.id)
def move_node(self, node1=None, node2=None):
"""
This function swaps the position of node1 with the position of node2
"""
#swapping two nodes!
logger.debug('Swapping two random nodes!: %s, %s', node1.id, node2.id)
# Swapping next nodes
logger.debug('Swapping next node')
tmp = copy.deepcopy(node1.next)
self._update_node(node=node1,
prev=node1.prev,
next=node2.next)
#Todo: Check if tmp changes or is stored as a copy
self._update_node(node=node2,
prev=node2.prev,
next=tmp)
if node1.next is not None:
logger.debug('Connect the next node to node 1')
node_next = self.get_node_from_node_id(node1.next)
self._update_node(node=node_next,
prev=node1.id,
next=node_next.next)
if node2.next is not None:
logger.debug('Connect the next node to node 2')
node_next = self.get_node_from_node_id(node2.next)
self._update_node(node=node_next,
prev=node2.id,
next=node_next.next)
logger.debug('Swap prev nodes')
tmp = copy.deepcopy(node1.prev)
self._update_node(node=node1,
prev=node2.prev,
next=node1.next)
self._update_node(node=node2,
prev=tmp,
next=node2.next)
# Connect the node before node1 to node 1
if node1.prev is not None:
logger.debug('Connect the prev to node 1')
node_prev = self.get_node_from_node_id(node1.prev)
self._update_node(node=node_prev,
prev=node_prev.prev,
next=node1.id)
# Connect the node before node2 to node 2
if node2.prev is not None:
logger.debug('Connect the prev to node 2')
node_prev = self.get_node_from_node_id(node2.prev)
self._update_node(node=node_prev,
prev=node_prev.prev,
next=node2.id)
global_test_array = []
obfuscation = 0xaa
class trivial_test_node_class(object):
def __init__(self, prev, id, next, data):
print "initializing test class."
# self.stuff = [ ["first", 0, 1], ["second", 0, 1] ]
self.id = id
self.prev = prev
self.next = next
self.data = data
def something(self):
print "something"
def save(self):
id = self.id
global_test_array[id] = self
def __repr__(self):
# print self.prev, self.id, self.next, self.data
the_string = "%s(%r)\n" % (self.__class__, self.__dict__)
return the_string
class trivial_test_list_model_class(object):
def __init__(self):
print "initializing test class."
#self.stuff = [ ["first", 0, 1], ["second", 0, 1] ]
self.stuff = [ 0 for i in xrange(0,10) ]
data = 'a'
for i in xrange(1,10):
self.stuff[i] = trivial_test_node_class(i-1,i,i+1,data);
data += 'r' # talk like a pirate day
self.stuff[-1].next = 0 # use 0 as NULL id.
global global_test_array
global_test_array = self.stuff
def something(self):
print "something"
def get(self,id):
return self.stuff[id]
def __repr__(self):
# for i in xrange(1,10):
# print self.stuff[i]
the_string = "%s(%r)" % (self.__class__, self.__dict__)
return the_string
if __name__ == '__main__':
# test code that only gets run when this file is run directly,
# not when this file is imported from some other python script.
print "Hello, world."
trivial_test_model = trivial_test_list_model_class()
print trivial_test_model
testll = DoublyLinkedList( trivial_test_model, "data" )
left_node = trivial_test_model.get(3)
right_node = trivial_test_model.get(4)
testll.move_node( left_node, right_node )
print trivial_test_model
# recommended by http://wiki.python.org/moin/vim
# vim: tabstop=8 expandtab shiftwidth=4 softtabstop=4 :
Alright,
I have this program to sparse code in Newick Format, which extracts both a name, and a distance for use in a phylogenetic tree diagram.
What my problem is, in this branch of code, as the program reads through the newickNode function, it assigns the name and distance to the 'node' variable, then returns it back into the 'Node' class to be printed, but it seems to only print the first node 'A', and skips the other 3.
Is there anyway to finish the for loop in newickNode to read the other 3 nodes and print them accordingly with the first?
class Node:
def __init__(self, name, distance, parent=None):
self.name = name
self.distance = distance
self.children = []
self.parent = parent
def displayNode(self):
print "Name:",self.name,",Distance:",self.distance,",Children:",self.children,",Parent:",self.parent
def newickNode(newickString, parent=None):
String = newickString[1:-1].split(',')
for x in String:
splitString = x.split(':')
nodeName = splitString[0]
nodeDistance = float(splitString[1])
node = Node(nodeName, nodeDistance, parent)
return node
Node1 = newickNode('(A:0.1,B:0.2,C:0.3,D:0.4)')
Node1.displayNode()
Thanks!
You could make it a generator:
def newickNode(newickString, parent=None):
String = newickString[1:-1].split(',')
for x in String:
splitString = x.split(':')
nodeName = splitString[0]
nodeDistance = float(splitString[1])
node = Node(nodeName, nodeDistance, parent)
yield node
for node in newickNode('(A:0.1,B:0.2,C:0.3,D:0.4)'):
node.displayNode()
The generator will return one node at a time and pause within the function, and then resume when you want the next one.
Or just save them up and return them
def newickNode(newickString, parent=None):
String = newickString[1:-1].split(',')
nodes = []
for x in String:
splitString = x.split(':')
nodeName = splitString[0]
nodeDistance = float(splitString[1])
node = Node(nodeName, nodeDistance, parent)
nodes.append(node)
return nodes
Your newickNode() function should accumulate a list of nodes and return that, rather than returning the first node created. If you're going to do that, why have a loop to begin with?
def newickNodes(newickString, parent=None):
nodes = []
for node in newickString[1:-1].split(','):
nodeName, nodeDistance = node.split(':')
nodes.append(Node(nodeName, nodeDistance, parent))
return nodes
Alternatively, you could write it as a generator that yields the nodes one at a time. This would allow you to easily iterate over them or convert them to a list depending on your needs.
def newickNodes(newickString, parent=None):
for node in newickString[1:-1].split(','):
nodeName, nodeDistance = node.split(':')
yield Node(nodeName, nodeDistance, parent)
Also, from a object-oriented design POV, this should probably be a class method on your Node class named parseNewickString() or similar.
Alternatively, your newickNode() function could immediately call node.displayNode() on the new node each time through the loop.
To keep this more flexible - I would use pyparsing to process the Newick text and networkx so I had all the graph functionality I could desire - recommend to easy_install/pip those modules. It's also nice that someone has written a parser with node and tree creation already (although it looks like it lacks some features, it'll work for your case):
http://code.google.com/p/phylopy/source/browse/trunk/src/phylopy/newick.py?r=66
The first time through your for: loop, you return a node, which stops the function executing.
If you want to return a list of nodes, create the list at the top of the function, append to it each time through the loop, and return the list when you're done.
It may make more sense to move the loop outside of the newickNode function, and have that function only return a single node as its name suggests.
Starting some programming with python at school now, and I don't know how to proceed with this problem. Any thoughts?
Input consists of integer separated by line breaks. Your program should submit them in a linked list, traverse the linked list and print the highest number.
Something to take the first number, and do an action which says "if the next number is bigger, take that one, else, keep the current number, and head down the list and repeat"
Then when it gets to the end of the list, it prints the value it has.
from sys import stdin
class Kubbe:
vekt = None
neste = None
def __init__(self, vekt):
self.vekt = vekt
self.neste = None
def spor(kubbe):
# WRITE YOUR CODE HERE
# Creates linked list
forste = None
siste = None
for linje in stdin:
forrige_siste = siste
siste = Kubbe(int(linje))
if forste == None:
forste = siste
else:
forrige_siste.neste = siste
# Calls the solution function and prints the result
print spor(forste)
Input: example
54
37
100
123
1
54
Required output
123
"Linked lists" are rarely used in Python -- normally, one uses just list, the Python built-in list, which is actually more of a "dynamic vector". So, it's peculiar to see a linked list specified as part of the exercise's constraints.
But the main point is, the code you're showing is already creating a linked list -- the head is at forste, and, for each node, the next-node pointer at .neste, the payload at .vekt. So, presumably, that's not what you're asking about, no matter the text of your question.
The simple way to loop through your linked list once you have fully constructed it (i.e., at the end of the current code for spor) is
current = forste
while current is not None:
...process current.vekt...
current = current.neste
In your case, the logic for the "process" part is of course, as your Q's text already says:
if current.vekt > themax:
themax = current.vekt
The only subtlety is, you need to initially set themax, before this while loop to "the lowest possible number"; in recent versions of Python, "minus infinity" is reliably recorded and compared (though only as a float, it still compares correctly to ints), so
themax = float('-inf')
would work. More elegant might be to initially set the maximum to the first payload, avoiding messing with infinity.
Here's an answer based on your own code and language. Sorry if the new variable and function names do not translate well, as I don't speak Norwegian (Google Language Tools is my friend).
Comment: Like airplane Air Traffic Control the default language of most international programming forums such as StackOverflow is English. If you use it, you are likely to get quicker, better, and more answers -- and it probably makes the question and related answers useful to the largest number of other folks. Just my 2 øre... ;-)
from sys import stdin
class Kubbe:
vekt = None
neste = None
def __init__(self, vekt):
self.vekt = vekt
self.neste = None
def spor():
# WRITE YOUR CODE HERE
# Creates linked list
forste = None
siste = None
while True:
try:
linje = raw_input()
except EOFError:
break
forrige_siste = siste
siste = Kubbe(int(linje))
if forste == None:
forste = siste
else:
forrige_siste.neste = siste
return forste
def finne_maksimal(lenketliste):
storste = None
if lenketliste is not None:
storste = lenketliste.vekt
gjeldende = lenketliste.neste
while gjeldende is not None:
if gjeldende.vekt > storste:
storste = gjeldende.vekt
gjeldende = gjeldende.neste
return storste
lenketliste = spor()
storste = finne_maksimal(lenketliste)
if lenketliste is None:
print "tom liste"
else:
print "storste er", storste
There is a builtin function in Python called reduce, which traverses a list and "compresses" it with a given function. That is, if you have a list of five elements [a,b,c,d,e] and a function f, it will effectively do
temp = f(a,b)
temp = f( temp, c )
...
You should be able to use this to write a very neat solution.
If you want to be less abstract, you will need to iterate over each element of the list in turn, storing the greatest number so far in a variable. Change the variable only if the element you have reached is greater than the value of said variable.
This seems to work with your input (works in both python 2 and 3). Notice how max works with duck typing of Python!
This version works with Python3 also from file.
import sys
class Kubbe:
vekt = None
neste = None
def __init__(self, vekt):
self.vekt = vekt
self.neste = None
def spor():
# WRITE YOUR CODE HERE
# Creates linked list
forste = None
siste = None
while True:
linje = sys.stdin.readline().rstrip()
if not linje:
break
forrige_siste, siste = siste, Kubbe(int(linje))
if forste is None:
forste = siste
else:
forrige_siste.neste = siste
return forste
def traverse(linkedlist):
while linkedlist is not None:
yield linkedlist.vekt
linkedlist=linkedlist.neste
# Calls the solution function and prints the result
linkedlist=spor()
for item in traverse(linkedlist):
print(item)
# use builtin max:
print('Maximum is %i' % max(traverse(linkedlist)))
# if not allowed:
m = linkedlist.vekt
for item in traverse(linkedlist.neste):
if item > m: m = item
print(m)
The below code would work. The Node class represents the LinkedList Node. The LinkedList class defines the methods to add node at the end of the Linked List and find_max will traverse through the list and return the node with largest key.
class Node(object):
def __init__(self, key, next_node):
self.key = key
self.next_node = next_node
class LinkedList(object):
def __init__(self):
self.head = None
def append(self, key):
# Create a new Node
new_node = Node(key, None)
if (self.head == None):
self.head = new_node
else:
tmp = self.head
while(tmp.next_node != None):
tmp = tmp.next_node
tmp.next_node = new_node
def find_max(self):
tmp = self.head
max_num = 0
while(tmp != None):
if (tmp.key > max_num):
max_num = tmp.key
tmp = tmp.next_node
return max_num