remove nth node from end of list

my solution:

python

# 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]:
        pir = head
        while n and pir:
            pir = pir.next
            n -= 1
        if n > 0:
            return head
        dummy = ListNode(next=head)
        pre = dummy
        cur = head
        while pir:
            pre = pre.next
            cur = cur.next
            pir = pir.next
        pre.next = cur.next
        return dummy.next

Note that: return the dummy.next due to the header could be deleted.

Logics can be modified:

python

class Solution:
    def removeNthFromEnd(self, head: Optional[ListNode], n: int) -> Optional[ListNode]:
        # 由于可能会删除链表头部，用哨兵节点简化代码
        left = right = dummy = ListNode(next=head)
        for _ in range(n):
            right = right.next  # 右指针先向右走 n 步
        while right.next:
            left = left.next
            right = right.next  # 左右指针一起走
        left.next = left.next.next  # 左指针的下一个节点就是倒数第 n 个节点
        return dummy.next

作者：灵茶山艾府
链接：https://leetcode.cn/problems/remove-nth-node-from-end-of-list/solutions/2004057/ru-he-shan-chu-jie-dian-liu-fen-zhong-ga-xpfs/
来源：力扣（LeetCode）
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