【LeetCode每天一题】Remove Nth Node From End of List(移除链表倒数第N个节点)

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.

Follow up:                  Could you do this in one pass?

思路

　　我们采用两个指针和哨兵模式来解决这个问题。 一开始先将快指针向前移动N位，然后和慢指针一起移动，直到指针指向最后一位结束。然后将漫指针的next指针重新赋值，就可以将对应节点删除。

　　时间复杂度为O(n)， 空间复杂度为O(1)。

图示步骤

解决代码

 1 # Definition for singly-linked list.
 2 # class ListNode(object):
 3 # def __init__(self, x):
 4 # self.val = x
 5 # self.next = None
 6 
 7 class Solution(object):  8     def removeNthFromEnd(self, head, n):  9         """
10  :type head: ListNode 11  :type n: int 12  :rtype: ListNode 13         """
14         if not head or n == 0: 15             return head 16         res = second = first = ListNode(0) # 构建哨兵节点 17         first.next = head 18         while n > 0: # 移动快指针 19             first = first.next 20             n -= 1
21         while first.next: # 移动快慢指针 22             second = second.next 23             first = first.next 24         second.next = second.next.next # 删除指定节点 25         return res.next