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19. Remove Nth Node From End Of List

Linked List Two Pointers

Problem - 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 <= 30
  • 0 <= Node.val <= 100
  • 1 <= n <= sz

 

Follow up: Could you do this in one pass?

Solutions

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# 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]:
        temp = ListNode(0, head)
        last = curr = temp

        for i in range(n + 1):
            curr = curr.next
        while curr:
            curr = curr.next
            last = last.next

        last.next = last.next.next

        return temp.next #head[n - 1].next = head[n].next

Submission Stats:

  • Runtime: 0 ms (100.00%)
  • Memory: 17.8 MB (54.84%)