【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

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　　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

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解决代码

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 # 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
         """
         if not head or n == 0:
             return head
         res = second = first = ListNode(0)       # 构建哨兵节点
         first.next = head
         while n > 0:                              # 移动快指针
             first = first.next
             n -= 1
         while first.next:                     # 移动快慢指针
             second = second.next
             first = first.next
         second.next = second.next.next         # 删除指定节点
         return res.next