Today I Learned: Remove Kth Node from the End of a Single Linked List
Anzhari Purnomo
Posted on September 14, 2020
Problem Statement
Write a function that takes in the head of a single linked list and integer k which represents the kth node to remove from the end of the linked list.
Sample Input
head = 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 9
k = 4
Sample Result
# node 6 is removed
0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 7 -> 8 -> 9
Code #1
class LinkedList:
def __init__(self, value):
self.value = value
self.next = None
def remove_kth_node_from_end(head, k):
ptr_1 = head
ptr_2 = head
ctr = 0
# Move ptr_2
while ctr < k:
# Return early if k is longer than list's length
if ctr < k and ptr_2 is None:
return
ptr_2 = ptr_2.next
ctr += 1
# Check if removing the first element on the list
if ptr_2 is None:
head.value = head.next.value
head.next = head.next.next
return head
while ptr_2.next is not None:
ptr_1 = ptr_1.next
ptr_2 = ptr_2.next
# Remove kth node from end
ptr_1.next = ptr_1.next.next
return head
Notes
- Since this is a single linked list, we cannot traverse to the end of the list and then trace back the kth node.
- Utilize two pointers to track the beginning and the end node of the linked list.
Credits
- Algoexpert for the problem statement
- Chris Leipelt.
💖 💪 🙅 🚩
Anzhari Purnomo
Posted on September 14, 2020
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