[Leetcode] 0019. Remove Nth Node From End of List

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Medium

Given the head of a linked list, remove the n ^th 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?

Table of contents

Python

				
					# time complexity: O(n) # space complexity: O(1) from typing import Optional 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]: total = 0 countNode = resNode = head while countNode: total += 1 countNode = countNode.next delIdx = total - n - 1 if delIdx < 0: head = head.next return head currentIdx = 0 while resNode: if currentIdx == delIdx: if resNode.next.next: resNode.next = resNode.next.next else: resNode.next = None return head else : resNode = resNode.next currentIdx += 1 return head root = ListNode(1) root.next = ListNode(2) root.next.next = ListNode(3) root.next.next.next = ListNode(4) root. next.next.next.next = ListNode(5) print(Solution().removeNthFromEnd(root, 2))

Python

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