题目描述:
中文:
给定一个链表,删除链表的倒数第 n 个节点,并且返回链表的头结点。
示例:
给定一个链表: 1->2->3->4->5, 和 n = 2.
当删除了倒数第二个节点后,链表变为 1->2->3->5.
说明:
给定的 n 保证是有效的。
进阶:
你能尝试使用一趟扫描实现吗?
英文:
Given a linked list, remove the n-th node from the end of list and return its head.
Example:
Given linked list: 1->2->3->4->5, and n = 2.
After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Follow up:
Could you do this in one pass?
# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None class Solution(object):
def removeNthFromEnd(self, head, n):
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
dummy = ListNode(0)
dummy.next = head
p1=p2=dummy
for i in range(n):
p1 = p1.next
while p1.next:
p1=p1.next
p2=p2.next
p2.next = p2.next.next
return dummy.next
题目来源:力扣