Description
Given the head of a linked list, remove the nth node from the end of the list and return its head.
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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]
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Constraints:
- The number of nodes in the list is sz.
- 1 <= sz <= 30
- 0 <= Node.val <= 100
- 1 <= n <= sz
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Follow up: Could you do this in one pass?
Solution
Python3
# 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]:
        res = curr = ListNode(-1, head)
        
        for _ in range(n):
            head = head.next
 
        while head and curr:
            head = head.next
            curr = curr.next
        
        curr.next = curr.next.next
 
        return res.next