Remove Nth Node From End of List
Given a linked list, remove the nth node from the end of list and return its head.
For 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.
Try to do this in one pass.
分析:
构建一个尺子,尺子长度为n,尺子顺着链表遍历,当尺子尾到达链表尾的时候,尺子头部记录要删除的结点,One pass算法。
代码:
class Solution(object):
def removeNthFromEnd(self, head, n):
"""
:type head: ListNode
:type n: int
:rtype: ListNode
"""
# 构建一个长度为n的尺子
rulertail = head
rulerhead = head
rulerheadpre = head
flag = True
for i in range(n - 1):
rulertail = rulertail.next
while rulertail.next is not None:
rulertail = rulertail.next
rulerhead = rulerhead.next
if flag:
flag = False
else:
rulerheadpre = rulerheadpre.next
# 删除倒数第n个元素
if rulerhead == rulertail:
if head == rulerhead:
return None
rulerheadpre.next = None
else:
rulerhead.val = rulerhead.next.val
rulerhead.next = rulerhead.next.next
return head