Problem

Given the head of a linked list, remove the nᵗʰ node from the end of the list and return its head.

https://leetcode.com/problems/remove-nth-node-from-end-of-list/

Example 1:

case1

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

Test Cases

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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]:
solution_test.py
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import pytest

from node import ListNode
from solution import Solution


@pytest.mark.parametrize('values, n, expected', [
([1,2,3,4,5], 2, [1,2,3,5]),
([1], 1, []),
([1,2], 1, [1]),

([1,2,3,4,5], 1, [1,2,3,4]),
([1,2,3,4,5], 5, [2,3,4,5]),
])
class Test:
def test_solution(self, values, n, expected):
sol = Solution()
head = self._build_list(values)
res_head = sol.removeNthFromEnd(head, n)
result = self._format(res_head)
assert result == expected

def _build_list(self, values: list[int]) -> ListNode | None:
root = ListNode()
node = root
for v in values:
node.next = node = ListNode(v)

return root.next

def _format(self, head: ListNode | None) -> list[int]:
values = []
while head is not None:
values.append(head.val)
head = head.next

return values

Thoughts

两个指针 p1、p2 都指向链表头,让 p2 先走 n 步,然后与 p1 一起往前走。每次让 p2 先走,没到的话再让 p1 走,这样当 p2 走到头的时候,p1 指向倒数第 n 个节点的前一个。

给链表前边加一个虚拟节点,指向 head,在处理边界的时候会方便很多。

Code

solution.py
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from typing import Optional

from node import ListNode

# 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]:
root = ListNode(None, head)
p1 = p2 = root
for _ in range(n):
p2 = p2.next

while True:
p2 = p2.next
if p2 is None:
break
p1 = p1.next

p1.next = p1.next.next

return root.next