删除链表的倒数第 N 个结点 - 力扣(LeetCode) 给你一个链表,删除链表的倒数第 n 个结点,并且返回链表的头结点。...示例 1:
输入:head = [1,2,3,4,5], n = 2
输出:[1,2,3,5]
示例 2:
输入:head = [1], n = 1
输出:[]
示例 3:
输入:head...= [1,2], n = 1
输出:[1]
提示:
链表中结点的数目为 sz
1 <= sz <= 30
0 <= Node.val <= 100
1 n <= sz
进阶: 你能尝试使用一趟扫描实现吗...先让 right 快指针 走 n 步
for i in range(n):
right = right.next
# 2....= None:
right = right.next
left = left.next
# 跳过中间的节点(即删除倒数第n个节点)