本文共 907 字，大约阅读时间需要 3 分钟。

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.

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def removeNthFromEnd(self, head, n):        """        :type head: ListNode        :type n: int        :rtype: ListNode        """        fast=head        slow=head                while n>0:            fast = fast.next            n -= 1                    if fast is None:            return head.next                while fast.next:            slow=slow.next            fast=fast.next                    slow.next = slow.next.next        slow=slow.next                return head

这道题想在O(n)时间解决，需要用到两个指针，分别设为fast和slow，中间固定距离n。

转载地址：http://ybrbb.baihongyu.com/