Remove Nth Node From End of List
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Problem Classification
Language Agnostic Coding Drills
Understand a Basic Class in Python: A Python class creates a new type where objects are instances of the class. Here
Solution
is the class.Understanding Linked Lists: Linked lists are data structures composed of nodes, each containing a value and a reference to the next node in the list. In Python, linked lists are not built-in data structures but are fundamental to this problem.
Linked List Traversal: Understanding how to traverse a linked list is crucial for this problem. This is done by setting a variable to the head node and then iteratively following the
next
references until you reach a node wherenext
isNone
.Defining and Using Functions Within Classes (Methods): Methods are functions that are defined within a class. In this case, the method
removeNthFromEnd
is defined within theSolution
class.The Concept of Fast and Slow Pointers: This is a common strategy used for traversing linked lists. By moving one pointer faster than the other, you can solve a variety of problems. Here,
fast
andslow
pointers are used to identify the node to be deleted.Loops and Conditionals: Understanding how loops (like
for
andwhile
loops) and conditionals (if
statements) work in Python is crucial to understanding this code.Null Checking: Checking whether a variable is
None
is very important in linked list problems to prevent accessingnext
of aNone
object.Modifying a Linked List: Finally, this problem involves modifying the linked list by skipping over the
n
th node from the end, which is accomplished by adjusting thenext
pointer of the node before it to point to the node after it.
Targeted Drills in Python
Python Class Creation
Create a class
Test
and instantiate an object from it. Print a message in the constructor.1 2 3 4 5
class Test: def __init__(self): print("Test class object created") test_object = Test()
Linked List Creation
Create a simple linked list structure. Make a Node class that has an integer value and a “next” field. Then, create a LinkedList class and initialize a list with some numbers.
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
class Node: def __init__(self, data): self.data = data self.next = None class LinkedList: def __init__(self): self.head = None ll = LinkedList() ll.head = Node(1) second = Node(2) third = Node(3) ll.head.next = second second.next = third
Linked List Traversal
Traverse the linked list created above and print the values of each node.
1 2 3 4
current = ll.head while current: print(current.data) current = current.next
Defining and Using Functions Within Classes (Methods)
Define a
print_list
method within theLinkedList
class that does what we did in drill 3. Then call this method on an instance ofLinkedList
.1 2 3 4 5 6 7 8 9 10
class LinkedList: #...previous code... def print_list(self): current = self.head while current: print(current.data) current = current.next ll.print_list()
The Concept of Fast and Slow Pointers
Implement a function that takes a linked list and returns the middle element of the linked list using the fast and slow pointer strategy.
1 2 3 4 5 6 7 8
def find_middle(ll): slow = fast = ll.head while fast and fast.next: slow = slow.next fast = fast.next.next return slow.data print(find_middle(ll))
Loops and Conditionals
Implement a function that takes a linked list and an integer
n
and prints then
th node in the list. You’ll need to include a conditional to check whethern
is greater than the length of the list.1 2 3 4 5 6 7 8 9
def print_nth_node(ll, n): count = 0 current = ll.head while current: if count == n: print(current.data) count += 1 current = current.next print_nth_node(ll, 2)
Null Checking
Modify the function
print_nth_node
so that it returns “Index out of range” ifn
is greater than the length of the list.1 2 3 4 5 6 7 8 9 10 11
def print_nth_node(ll, n): count = 0 current = ll.head while current: if count == n: print(current.data) return count += 1 current = current.next print("Index out of range") print_nth_node(ll, 5)
Modifying a Linked List
Implement a function that removes the
n
th node from the end of the list. This will involve reassigning thenext
pointer of the (n+1
)th node from the end to the (n-1
)th node from the end.1 2
def remove_nth_from_end(ll, n): dummy =
Node(0) dummy.next = ll.head slow = fast = dummy for _ in range(n+1): fast = fast.next while fast: slow = slow.next fast = fast.next slow.next = slow.next.next ll.head = dummy.next
remove_nth_from_end(ll, 2)
ll.print_list()
```
Remember to always test your functions with different inputs to ensure they work as expected.