19. Remove Nth Node From End of List [easy] (Python)

本文链接：https://blog.youkuaiyun.com/coder_orz/article/details/51691267

题目链接

https://leetcode.com/problems/remove-nth-node-from-end-of-list/

题目原文

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.

题目翻译

给定一个链表，删除从尾部开始数的第n个节点，并返回链表头部。
比如：给定链表1->2->3->4->5，以及 n = 2 ，删除节点后，链表变成 1->2->3->5。
注意：给定的n总是有效的；尽量在一遍遍历的情况下完成这道题。

思路方法

思路一

用两个指针，一个指针p从前到后扫描整个链表，一个指针q慢指针p的步数为n+1，那么当p指向尾部的Null时，指针q恰好指向要删除节点的前一个节点。由于可能删除头部节点，伪装一个新的头部方便操作。

代码

# 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
        """
        new_head = ListNode(0)
        new_head.next = head
        fast = slow = new_head
        for i in range(n+1):
            fast = fast.next
        while fast:
            fast = fast.next
            slow = slow.next
        slow.next = slow.next.next
        return new_head.next

思路二

题目并没有说不能修改节点的值，一个比较tricky的做法是：将倒数第n个节点前的节点的值依次后移，最后返回head.next即为所求。用递归实现了“倒着数”的操作。

代码

# 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
        """
        def getIndex(node):
            if not node:
                return 0
            index = getIndex(node.next) + 1
            if index > n:
                node.next.val = node.val
            return index
        getIndex(head)
        return head.next

思路三

类似思路二，不过这次不再修改节点值，而是真正的删除倒数第n个节点。

代码

# 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
        """
        def remove(node):
            if not node:
                return 0, node
            index, node.next = remove(node.next)
            next_node = node if n != index+1 else node.next
            return index+1, next_node
        ind, new_head = remove(head)
        return new_head