【Description】
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.
【Idea】
-
第一种解法：
遍历该逆向列表，并用list存储。del nums[length-n] 删除指定下标元素，再重新遍历list，创建新链表。提交结果如图，空间占用较高。
一.png -
第二种解法：
在原链表基本上，创建两个指针fast 和slow， 对链表进行时间复杂度为O（n）的遍历，中fast所指节点的位置比slow 快n。遍历至fast.next==None,将slow.next.next赋给slow.next。最后返回head。提交结果如下（空间并没有低。。）：
二.png
【Solution】
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, x):
# self.val = x
# self.next = None
# solution 1
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
temp = head
nums = []
while temp:
nums.append(temp.val)
temp = temp.next
length = len(nums)
if n > length:
return
if length == 1 and n == 1:
return []
del nums[length-n]
head = ListNode(None)
prev = head
for i in nums:
curr = ListNode(i)
prev.next = curr
prev = prev.next
return head.next
# solution 2
class Solution:
def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode:
fast = slow = head
for i in range(n):
fast = fast.next
if fast == None:
return head.next
while fast.next:
slow = slow.next
fast = fast.next
slow.next = slow.next.next
return head
