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。(python)
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?
题目链接:https://leetcode-cn.com/problems/remove-nth-node-from-end-of-list/description/
答案——
方法一:
# 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
"""
be = ListNode(None) #在链表最前面添加一个be空节点
be.next = head
p = be
p1 = p
p2 = p
p3 = p
length = 0
while p.next <> None: #获得链表长度
length = length + 1
p = p.next
if n == length: #删除第一个节点
head = head.next
elif n == 1: #删除最后一个节点
for k in range(length -1):
p3 = p3.next
p3.next = None
else: #删除中间的节点
for i in range(length - n):
p1 = p1.next
for j in range(length - n + 2):
p2 = p2.next
p1.next = p2
return head
方法二(进阶):
# 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
"""
be = ListNode(0)
be.next = head
p1 = be
p2 = be
for i in range(n + 1):
p1 = p1.next
while p1 <> None:
p1 = p1.next
p2 = p2.next
p2.next = p2.next.next
return be.next
使用两个指针,保证两指针的距离为n,当后面的指针到末尾时,前面的指针在倒数第n个位置上。