It’s been a while since I have written something here. A lot of work is going on and haven’t got much time tbh. So, here’s a piece of recursive code to delete the N-th node from the end of a singly linked list.
The problem statement (from leetcode) is as follows:
Given the head of a linked list, remove the nth node from the end of the list and return its head.
Example 1:
Input: head = [1,2,3,4,5], n = 2 Output: [1,2,3,5]
Example 2:
Input: head = [1], n = 1 Output: []
Example 3:
Input: head = [1,2], n = 1 Output: [1]
Constraints:
- The number of nodes in the list is
sz. 1 <= sz <= 300 <= Node.val <= 1001 <= n <= sz
Well, an array of solutions are possible for this problem, but I have tried the recursive one which passed all the test cases. Given below is the solution class, the driver class for the same can be found here link.
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
curr_node = head
curr_pos = 0
def mark_node(curr, n):
if not curr.next:
curr_pos = 2
if n == 1:
return (curr_pos, False)
return (curr_pos, True)
curr_pos, flag = mark_node(curr.next, n)
if curr_pos == n:
return curr_pos + 1, False
if not flag:
curr.next = curr.next.next
return curr_pos+1, True
_, flag = mark_node(curr_node, n)
if not flag:
head = head.next
return head
Well let me know if this can be optimised further.
Thanks
Technical Dopamine 
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