19. Remove Nth Node From End of List
Medium
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
Follow up: Could you do this in one pass?
# 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: Optional[ListNode], n: int) -> Optional[ListNode]:
# size = 0
# node = head
# while node:
# size+=1
# node = node.next
# node = head
# target_node_position = size - n
# # head
# if target_node_position == 0:
# return head.next
# target_node_position -=1
# # head is removed, now the rest of them
# for i in range(target_node_position+1):
# if i == target_node_position:
# node.next = node.next.next
# node = node.next
# return head
# two pointer approach
current_node = head
for i in range(n):
current_node=current_node.next
if current_node is None:
return head.next
node_before_remove = head
while current_node.next is not None:
current_node = current_node.next
node_before_remove = node_before_remove.next
node_before_remove.next = node_before_remove.next.next
return head