Delete a Node at a Given Position in a Linked List
Implement deleteAtPosition
Given the
head of a singly linked list and a 0-indexed position, delete the node at that index and return the head of the resulting list.
position = 0 is exactly the delete-at-the-beginning case. For any other position, the node being deleted is never visited directly — only the node right before it matters, since deleting means redirecting that node's next pointer straight past the target.
Example 1:
Input: head = [1,2,3,4,5], position = 2
Output: [1,2,4,5]
Example 2:
Input: head = [9], position = 0
Output: []
Example 3:
Input: head = [10,20,30], position = 0
Output: [20,30]
+ 5 hidden test cases run on Submit.
Constraints:
- ●
1 ≤ number of nodes in head ≤ 10⁴ - ●
-10⁹ ≤ node value ≤ 10⁹ - ●
0 ≤ position < length of head — position is 0-indexed, so position 0 deletes the head
head =
[1, 2, 3, 4, 5]
position =
2