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[9/28] 19. Remove Nth Node From End of List

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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?

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Runtime: 60 ms, faster than 41.68% of Python3 online submissions for Remove Nth Node From End of List.
Memory Usage: 14 MB, less than 20.71% of Python3 online submissions for Remove Nth Node From End of List.
# 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]:
        count = 0
        curr = head
        while curr:
            count += 1
            curr = curr.next
        
        dummy = ListNode(0, head)
        left, right = dummy, head.next
        LEFT, RIGHT = count - n, count - n + 2
        

        for i in range(LEFT):
            right = right.next
            left = left.next
        
        left.next = right
        
        return dummy.next
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