Remove Nth to Last Node | Python
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class Node:
def __init__(self, value, next=None):
self.value = value
self.next = next
def __str__(self):
return str(self.value)
class LinkedList:
def __init__(self, *items):
self.head = None
self.tail = None
self.add(*items)
def add(self, *items):
for item in items:
self.addAtEnd(item)
def addAtEnd(self, value):
if not self.head:
self.head = self.tail = Node(value)
else:
curr = self.head
while curr.next:
curr = curr.next
curr.next = Node(value)
self.tail = curr.next
def elems(self):
curr = self.head
count = 0
while curr:
count += 1
curr = curr.next
return count
def __str__(self):
curr = self.head
res = []
while curr:
res.append(str(curr.value))
curr = curr.next
return "->".join(res)
def removeNthToLastNode(lst: LinkedList, n: int):
numOfItems = lst.elems()
if n > numOfItems:
raise Exception(f"Index out of Range. Expected: 1..{numOfItems}. Got: {n}")
prev = None
curr = lst.head
while curr and numOfItems != n:
prev = curr
curr = curr.next
numOfItems -= 1
if prev and curr:
prev.next = curr.next
elif curr and numOfItems == n:
lst.head = lst.head.next
return lst
tests = [
([1, 2, 3], 1, '1->2'),
([1, 2, 3], 2, '1->3'),
([1, 2, 3], 3, '2->3'),
([4, 4, 7, 1, 5, 6], 0, '4->4->7->1->5->6'),
([4, 4, 7, 1, 5, 6], 1, '4->4->7->1->5'),
([4, 4, 7, 1, 5, 6], 6, '4->7->1->5->6'),
([4, 4, 7, 1, 5, 6], 2, '4->4->7->1->6'),
([4, 4, 7, 1, 5, 6], 5, '4->7->1->5->6'),
]
for i, (a, n, expected) in enumerate(tests):
l = LinkedList(*a)
got = str(removeNthToLastNode(l, n))
if got == expected:
print(f"PASS")
else:
print(f"FAIL: expected '{expected}', got '{got}'")
Python
INFO